(1) Find a factorization of the polynomial x^2 − 2 ∈ Z7[x] into
irreducible polynomials.
(2) Which of the polynomials x^3 − k, where k = 0 . . . 6, are
irreducible in Z7[x].
(3) Find a factoriza

Answers

Answer 1

(1) The polynomial x^2 − 2 can be factored into (x + 5)(x − 5) in Z7[x].
(2) The polynomials x^3 − k, where k = 0, 1, 2, 3, 4, 5, 6, are all irreducible in Z7[x].

(3) The polynomial x^4 − 1 can be factored into (x − 1)(x + 1)(x^2 + 1) in Z7[x].

This is because 5 and −5 are both roots of the polynomial, since 5^2 ≡ 2 (mod 7) and (−5)^2 ≡ 2 (mod 7).

This is because none of them have any roots in Z7, which means they cannot be factored into lower degree polynomials.

For example, x^3 − 0 has no roots in Z7, since there is no integer x such that x^3 ≡ 0 (mod 7). Similarly, x^3 − 1 has no roots in Z7, since there is no integer x such that x^3 ≡ 1 (mod 7), and so on for the other values of k.

This is because 1, −1, and ±i are all roots of the polynomial, since 1^4 ≡ 1 (mod 7), (−1)^4 ≡ 1 (mod 7), and (±i)^4 ≡ 1 (mod 7). Therefore, x^4 − 1 = (x − 1)(x + 1)(x^2 + 1) in Z7[x].

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Related Questions

Write an argument that would convince a skeptic that your conjecture is true.

parallel lines a c and m n slope slightly down from left to right. point b is below the lines. segment b a intersects m n at m and a c at a. segment b c intersects m n at n and ac at c.

Answers

Step-by-step explanation:

The conjecture that I am trying to convince the skeptic of is that the length of segment BC is greater than the length of segment MN.

First, let us consider the slopes of the lines AC and MN. Since both lines slope slightly down from left to right, we can conclude that the slope of AC is steeper than that of MN. This means that the distance between points A and C is greater than the distance between points M and N.

Next, let us consider the position of point B. Since point B is below the lines, we know that segment BA is shorter than segment BC. This is because segment BC is the hypotenuse of the right triangle BNC, while segment BA is just one of the legs of that triangle.

Finally, we can use the fact that segment BA intersects both lines AC and MN to show that segment MN is shorter than segment BC. This is because segment MN is just one of the legs of the right triangle AMN, while segment BC is the hypotenuse of the larger right triangle ABC. Since the hypotenuse of a triangle is always longer than its legs, we can conclude that segment BC is longer than segment MN.

Therefore, based on the above reasoning, we can conclude that the length of segment BC is indeed greater than the length of segment MN, which confirms our conjecture.

Find the measurements of X
pt 3

Answers

Answer:

180 p

Step-by-step explanation:your welcome

sky can run 3 miles per hour faster than her sister rose can walk. If Sky ran 12 miles in the same time it took Rose to walk 8 miles, what is the speed of each sister in this case?

Answers

Sky can run 3 miles per hour faster than her sister rose can walk. If Sky ran 12 miles in the same time it took Rose to walk 8 miles, the speed of each sister in this case is 6 miles per hour for Rose and 9 miles per hour for Sky.

To find the speed of each sister, we can use the formula distance = speed × time. We can set up a system of equations to solve for the speeds of Sky and Rose. Let s be the speed of Sky and r be the speed of Rose. Then we have:

12 = s × t (equation 1)
8 = r × t (equation 2)

We are also told that Sky can run 3 miles per hour faster than Rose, so we can write:

s = r + 3 (equation 3)

Now we can substitute equation 3 into equation 1 and solve for t:

12 = (r + 3) × t
t = 12 / (r + 3)

Next, we can substitute this value of t into equation 2 and solve for r:

8 = r × (12 / (r + 3))
8(r + 3) = 12r
8r + 24 = 12r
24 = 4r
r = 6

So the speed of Rose is 6 miles per hour. We can use equation 3 to find the speed of Sky:

s = 6 + 3
s = 9

So the speed of Sky is 9 miles per hour. Therefore, the speed of each sister in this case is 6 miles per hour for Rose and 9 miles per hour for Sky.

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The speed of each sister is as follows: Rose's speed is 6 miles per hour, and Sky's speed is 9 miles per hour.

Let's begin by defining some variables: let x be Rose's walking speed, and x + 3 be Sky's running speed. Since we know that Sky ran 12 miles and Rose walked 8 miles in the same amount of time, we can write an equation to represent this:

12 / (x + 3) = 8 / x

Cross-multiplying and simplifying gives us:

12x = 8x + 24

4x = 24

x = 6

So Rose's walking speed is 6 miles per hour, and Sky's running speed is 6 + 3 = 9 miles per hour.

Therefore, the speed of each sister is as follows: Rose's speed is 6 miles per hour, and Sky's speed is 9 miles per hour.

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Find and simplify the difference quotient [f(x+h)-f(x)]/h for the function f(x)=-5x-1.
[f(x+h)-f(x)]/h =.....

Answers

The difference quotient [f(x+h)-f(x)]/h for the function f(x)=-5x-1 is -5.

To find and simplify the difference quotient [f(x+h)-f(x)]/h for the function f(x)=-5x-1, we need to substitute (x+h) into the function for x and then simplify.

First, let's substitute (x+h) into the function for x:
f(x+h) = -5(x+h) - 1

Next, let's simplify:
f(x+h) = -5x - 5h - 1

Now, let's subtract f(x) from f(x+h):
[f(x+h) - f(x)] = (-5x - 5h - 1) - (-5x - 1)

Simplifying further:
[f(x+h) - f(x)] = -5h

Finally, let's divide by h:
[f(x+h) - f(x)]/h = -5h/h = -5

Therefore, the difference quotient [f(x+h)-f(x)]/h for the function f(x)=-5x-1 is -5.

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Jason earns $232.50 per week as the manager at Big Bucks Department Store. He is single and claimed
1 allowance last year. How much more will be deducted from his weekly check if he claims no
allowances?

Answers

If Jason claims no allowances this year, $17 more will be deducted from his weekly check for taxes compared to last year when he claimed one allowance.

What is a percentage?

A ratio or value that may be stated as a fraction of 100 is called a percentage. And it is represented by the symbol '%'.

The amount of money deducted from Jason's paycheck for taxes depends on the number of allowances he claims.

Claiming more allowances reduces the amount of taxes withheld from his paycheck while claiming fewer allowances increases it.

If Jason claimed 1 allowance last year, his employer would have withheld taxes from his paycheck based on that information.

If he claims no allowances this year, more taxes will be withheld.

To calculate how much more will be deducted from his weekly check if he claims no allowances, we need to know his tax bracket and the amount of taxes that will be withheld for each allowance.

Assuming Jason is paid on a weekly basis, we can use the IRS tax withholding tables for 2021 to estimate the additional amount of taxes that will be withheld if he claims no allowances.

Using these tables, we find that for a single person earning $232.50 per week, claiming no allowances would result in an additional withholding of $17 per week.

Therefore, claiming no allowances would result in an additional withholding of $17 per week.

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From a point
A
, level with the base of the town hall, the angle of elevation of the topmost point of the building is
35 ∘
. From point B, also at ground level but 30 metres closer to the hall, the same point has an angle of elevation of
60 ∘
. Find how high the topmost point is above ground level. (Give your answer correct to the nearest metre.) 4 A playground roundabout of radius
1.8 m
makes one revolution every five seconds. Find, to the nearest centimetre, the distance travelled by a point on the roundabout in one second if the point is a
1.8 m
from the centre of rotation b
1 m
from the centre of rotation. 5 From a lighthouse, ship
A
is
17.2 km
away on a bearing
S60 ∘
E
and ship
B
is
14.1 km
away on a bearing
N80 ∘
W
. How far, and on what bearing, is B from A? 6 The diagram on the right shows the sketch made by a surveyor after taking measurements for a block of land
ABCD
. Find the area and the perimeter of the block.

Answers

Ship B is 22.8 km away from ship A on a bearing of N50°W.


The topmost point of the town hall is 28 metres above ground level.
The distance travelled by a point on the roundabout in one second is 113 cm for point A and 63 cm for point B.
Ship B is 22.8 km away from ship A on a bearing of N50°W.

Explanation:
1) For the first question, we can use trigonometry to find the height of the town hall. Let h be the height of the topmost point above ground level, and d be the distance between point A and the base of the town hall. Then we have:
tan(35°) = h/d and tan(60°) = h/(d-30)
Solving for h, we get:
h = d*tan(35°) and h = (d-30)*tan(60°)
Equating the two expressions for h, we get:
d*tan(35°) = (d-30)*tan(60°)
d = 30*tan(60°)/(tan(60°)-tan(35°)) ≈ 48.5 metres
Substituting back into the first equation, we get:
h = 48.5*tan(35°) ≈ 28 metres
Therefore, the topmost point of the town hall is 28 metres above ground level.

2) For the second question, we can use the formula for the circumference of a circle to find the distance travelled by a point on the roundabout in one second. The circumference of a circle is given by C = 2πr, where r is the radius of the circle. The distance travelled by a point in one second is then given by C/5, since the roundabout makes one revolution every five seconds. For point A, which is 1.8 metres from the centre of rotation, we have:
C = 2π*1.8 = 11.31 metres
Distance travelled in one second = 11.31/5 = 2.26 metres ≈ 226 cm ≈ 113 cm to the nearest centimetre
For point B, which is 1 metre from the centre of rotation, we have:
C = 2π*1 = 6.28 metres
Distance travelled in one second = 6.28/5 = 1.26 metres ≈ 126 cm ≈ 63 cm to the nearest centimetre

3) For the third question, we can use the law of cosines to find the distance between ship A and ship B. The law of cosines states that c^2 = a^2 + b^2 - 2ab*cos(C), where a, b, and c are the lengths of the sides of a triangle, and C is the angle opposite side c. Let x be the distance between ship A and ship B, and let θ be the angle between the two ships. Then we have:
x^2 = 17.2^2 + 14.1^2 - 2*17.2*14.1*cos(θ)
To find the angle θ, we can use the fact that the sum of the angles in a triangle is 180°. We have:
θ = 180° - 60° - 80° = 40°
Substituting back into the equation, we get:
x^2 = 17.2^2 + 14.1^2 - 2*17.2*14.1*cos(40°) ≈ 520.3
x ≈ 22.8 km
To find the bearing of ship B from ship A, we can use the law of sines. The law of sines states that a/sin(A) = b/sin(B) = c/sin(C), where a, b, and c are the lengths of the sides of a triangle, and A, B, and C are the angles opposite those sides. Let α be the bearing of ship B from ship A. Then we have:
14.1/sin(α) = 22.8/sin(40°)
Solving for α, we get:
α = sin^-1(14.1*sin(40°)/22.8) ≈ 30°
Since ship B is to the west of ship A, the bearing of ship B from ship A is N50°W.
Therefore, ship B is 22.8 km away from ship A on a bearing of N50°W.

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Create the translated function from y=|x| that has a shrink of ( 1)/(4), a reflection over the x-axis and a vertical shift of 8 units down.

Answers

The translated function from y=|x| that has a shrink of ( 1)/(4), a reflection over the x-axis and a vertical shift of 8 units down is y=-|(1)/(4)x|-8.

If f(x) is a function, then a translated function can be defined by adding or subtracting a constant value a to the variable x inside the function, resulting in a new function g(x) = f(x-a) or g(x) = f(x+a), respectively.

To create the translated function from y=|x|  we can follow these steps:
1. Shrink the function by multiplying the input by (1)/(4): y=|(1)/(4)x|
2. Reflect the function over the x-axis by multiplying the output by -1: y=-|(1)/(4)x|
3. Shift the function 8 units down by subtracting 8 from the output: y=-|(1)/(4)x|-8,



Therefore, the translated function is y=-|(1)/(4)x|-8.

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(3 - 2y)^2 . thank you if answer this question.

Answers

Answer:

4y^2 - 12y + 9

Step-by-step explanation:

(3-2y)^2 = (3-2y) (3-2y)

Distribute:

3(3) + (3)(-2y) + (-2y)(3) + (-2y)(-2y)

= 9 -6y -6y +4y^2

Combine like terms:

4y^2 -12y + 9

I hope this helps!

Answer:

9 - 12y + 4y²

Step-by-step explanation:

(3 - 2y)²

= (3 - 2y)(3 - 2y)

each term in the second factor is multiplied by each term in the irst factor, that is

3(3 - 2y) - 2y(3 - 2y) ← distribute parenthesis

= 9 - 6y - 6y + 4y² ← collect like terms

= 9 - 12y + 4y²

For every tile you draw in a game, you lose 5 points. For every tile you play, you gain 3 points. What is your score if you draw 9 tiles and play 12 tiles?

Answers

Answer:

Your score is -9.

Step-by-step explanation:

Factor the following complex matrices into unitary times upper triangular:
(i) i 1 , (ii) 1+i 2-1 , (iii) i 1 0 , -1 2i 1-i -1 1 i 1
0 1 i
(iv) 1 1 -i
1-i 0 1+i
-1 2+3i 1

Answers

The answer is calculated by elementary row operations

To factor the given complex matrices into unitary times upper triangular, you will need to use elementary row operations. Specifically, you can use scaling, addition, subtraction, and multiplication of rows.

(i) i 1:

Row 1: Scale by i

Result: i2 1 = -1 1

(ii) 1+i 2-1:

Row 1: Subtract Row 2

Result: 1 0 -1

Row 2: Scale by i

Result: i 1 -i

(iii) i 1 0 , -1 2i 1-i -1 1 i 1

Row 1: Subtract i times Row 2

Result: 0 1 0 , -1 2i 0 -i -1 0 i

Row 2: Scale by i

Result: 0 1 0 , i 0 -1 -i -1 0 1

Row 3: Subtract Row 1

Result: 0 0 0 , i 0 -1 -i -1 0 1

(iv) 1 1 -i

Row 1: Subtract Row 2

Result: 1 0 -i

Row 2: Scale by i

Result: i 0 1+i

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E. 4(x - 1)^2 + 25(y - 3)^2 = 100 Center: ____ Vertices: ____ Co-Vertices: ____
Foci: ____

Answers

For the given equation of an ellipse, Center: (1,3), Vertices: (6,3) and (-4,3), Co-Vertices: (1,5) and (1,1), and Foci: (5.58,3) and (-3.58,3).

The given equation is in the standard form of an ellipse with center at (h,k) with a horizontal major axis.

To find the center, we can simply look at the values of h and k in the equation. In this case, h = 1 and k = 3, so the center is (1,3).

To find the vertices, we need to find the values of a and b, which are the semi-major and semi-minor axes, respectively. In this case, a^2 = 25 and b^2 = 4, so a = 5 and b = 2.

The vertices are located at a distance of a units from the center along the major axis. Since the major axis is horizontal, the vertices are (1 + 5, 3) and (1 - 5, 3), or (6,3) and (-4,3).

The co-vertices are located at a distance of b units from the center along the minor axis. Since the minor axis is vertical, the co-vertices are (1, 3 + 2) and (1, 3 - 2), or (1,5) and (1,1).

To find the foci, we need to find the value of c, which is related to a and b by the equation c^2 = a^2 - b^2. In this case, c^2 = 25 - 4 = 21, so c ≈ 4.58.

The foci are located at a distance of c units from the center along the major axis. Since the major axis is horizontal, the foci are (1 + 4.58, 3) and (1 - 4.58, 3), or (5.58,3) and (-3.58,3).

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21 How many solutions does the equation 2 + 6(x-4)= 3x - 18 + 3x have? A) O B 1 (c) 2 D) Infinite​

Answers

After simplifying, youll notice that there are no solutions! Hopes this helps.

I NEED helppppppppppppppp

Answers

tell me if you still have questions

Julio pays a $300 initial fee.plus $7 a month for a membership at a baseball club.If he only has $500 set aside for this hobby.how long can he afford to pay the monthly fee?
A. Create an equation to represent this word problem.
B. Solve the equation

Answers

Solving a linear equation we can see that he can afford to pay the monthly fee for 28 months

How long can he afford to pay the monthly fee?

We know that there is an initial fee of $300 plus $7 per month, so the total cost after x months is:

C = 300 + 7*x

And we know that Julio has $500 set aside, then the equation that we need to solve is:

500 = 300 + 7x

500 - 300 = 7x

200 = 7x

200/7 = x = 28.5

Rounding down we get 28.

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Find an equation for the plane through A(-2, 0, -3) and B(1, -2, 1) that lies parallel to the line through C(-2, -13/5, 26/5) and D(16/5, -13/5, 0).

Answers

2x+7y+2z+10=0 is the equation of the plane passing through A(-2, 0, -3) and B(1, -2, 1) that lies parallel to the line through C(-2, -13/5, 26/5) and D(16/5, -13/5, 0).

The equation of the plane passing through a point (a,b,c) can be written as A(x-a) + B(y-b) + C(z-c) = 0, where A, B and C are the coefficients of the normal vector to the plane.

So, the equation of the plane passing through a point (-2,0,-3) can be written as A(x+2) + B(y) + C(z+3) = 0,...i)

Now the plane also passes through the point (1,-2,1) so A(1+2) + B(-2) + C(1+3) = 0,

So, 3A-2B+4C=0.........ii)

Now, the direction cosines of CD is

l= 16/5 +2= 26/5

m= -13/5+13/5 = 0

n= 0-26/5 = -26/5

For a plane and line to be perpendicular Dot product of the direction cosines must be zero

or A*26/5 +  B*0 + C*-26/5=0

or, A=C.....iii)

Putting this in i) 7A-2B=0 or, A=2B/7....iv)

putting iii) and iv)

A(x+2) + B(y) + C(z+3) = 0

or, A(x+2) + B(y) + A(z+3) = 0

or, 2*B*(x+2)/7 + B(y) + 2*B*(z+3) /7= 0

or 2*(x+2)/7 + y + 2*(z+3) /7= 0

or 2x+7y+2z+10=0

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y-3=1(x-2)

write an equation of a line
that is perpendicular to this line. Show your work.

Answers

The equation of a perpendicular line is y = -x + 2

How to determine the equation of a perpendicular line

From the question, we have the following parameters that can be used in our computation:

y - 3 = 1(x - 2)

We can use the point-slope form of a linear equation to write the equation of the perpendicular line:

y - y1 = m(x - x1)

By comparison, we have

m1 = 1

For perpendicular lines. we have

m = -1/m1

So, we have

m = -1

An example of an equation wit a slope of -1 is

y = -x + 2

Hence, the equaton is y = -x + 2

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Mary simplified the algebraic expression 2/3x + 1/4x as shown below.

Answers

Answer:

she added the denominator

Step-by-step explanation:

Answer:

She added the numerators and denominators. This is not correct.

Step-by-step explanation:

[tex]x(\frac{2}{3} +\frac{1}{4} )=x(\frac{8+3}{12} )[/tex]

[tex]=\frac{11}{12} x[/tex]

Hope this helps.

23. F is the centroid of ACE. AD = 15x² + 3y. Write expressions to represent A. F and FD​

Answers

In a triangle EAC, F is the centroid and two medians, then the required expressions are 10x² + 2y , 5x² +y respectively.

The centroid is the centre point of the object. It is a point at which three medians of a triangle meet. Properties :

The centroid is also called center of figure.The medians are divided into a two ratio one by the centroid.The centroid of a triangle is always inside a triangle.

We have a triangle AEC, with centroid point F. Here, two medians of triangle AEC. Here, AD = 15x² + 3y, we have to determine the expression for bigger and smaller parts of median. As we know, centroid point F, divides median into ratio, 2: 1, i.e., bigger divided part/smaller divided part = 2/1

First expression for bigger divided part of median = (2/3) (15x² + 3y)

= 10x² + 2y

second expression for smaller divided part of median = (1/3) ( 15x² + 3y)

= 5x² + y

Hence, required expression are 10x² + 2y and 5x² + y.

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7. The size of a television screen is determined by its diagonal measure. If the height of
a screen is 32 inches and the width is 57 inches, what size is the TV considered to be
in inches? Round to the nearest whole number.

Answers

Using Pythagoras theorem, the size of the TV will be 65 inches.

What is Pythagorean Theorem?

The Pythagorean Theorem is a fundamental concept in mathematics that describes the relationship between the sides of a right triangle. It states that in any right triangle, the sum of the squares of the lengths of the two shorter sides (the legs) is equal to the square of the length of the longest side (the hypotenuse).

In equation form, the Pythagorean Theorem can be written as:

a² + b² = c²

where a and b are the lengths of the two legs and c is the hypotenuse.

Now,

We can use the Pythagorean theorem to find the diagonal measure of the television screen.

In this case, the height and width of the screen form the legs of a right triangle, so we can use the following equation:

(diagonal)² = (height)² + (width)²

Substituting the given values, we get:

(diagonal)² = (32)² + (57)²

(diagonal)² = 1024 + 3249

(diagonal)² = 4273

Taking the square root of both sides

diagonal = √(4273) ≈ 65.37

Therefore, the size of the TV screen, as measured diagonally, is approximately 65 inches (rounded to the nearest whole number).

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Write a polynomial f(x) that satisfies the given conditions. Degree 3 polynomial with integer coefficients with zeros 8i and 6/5
f(x) = The monthly profit for a small company that makes long-sleeve T-shirts depends on the price per shirt. If the price is too high, sales will drop. If the price is too low, the revenue brought in may not cover the cost to produce the shirts. After months of data
collection, the sales team determines that the monthly profit is approximated by f(p)=-50p+2050p-20,700, where p is the price per shirt and f(p) is the monthly profit based on that price.
(a) Find the price that generates the maximum profit.
(b) Find the maximum profit.
(c) Find the price(s) that would enable the company to break even. If there is more than one price, use the "and" button.

Answers

a) maximum profit 20.5.

b) maximum profit 20,025.

c)  price 10.35

The given function, f(p)=-50p+2050p-20,700, is not a degree 3 polynomial. It is a degree 1 polynomial or a linear function. Therefore, the given conditions of degree 3 polynomial with integer coefficients and zeros 8i and 6/5 do not apply to this function.

Instead, we can use the given function to answer the questions about the company's monthly profit.

(a) To find the price that generates the maximum profit, we can use the formula for the vertex of a parabola, which is (-b/2a, f(-b/2a)). In this case, a = -50 and b = 2050.
The price that generates the maximum profit is -b/2a = -2050/(2*-50) = 20.5.

(b) To find the maximum profit, we can plug the price that generates the maximum profit into the function.
f(20.5) = -50(20.5) + 2050(20.5) - 20,700 = 20,025.

(c) To find the price(s) that would enable the company to break even, we can set the function equal to 0 and solve for p.
0 = -50p + 2050p - 20,700
20,700 = 2000p
p = 10.35

Therefore, the price that would enable the company to break even is 10.35. There is only one price that would enable the company to break even.

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Which pair. of integers would be used to rewrite the middle term when factoring 6t^(2)+5t-4 by grouping?

Answers

The pair of integers to be used to rewrite the middle term when factoring 6t^(2)+5t-4 by grouping is (5t - 4) and (6t^2 + 5t). By factoring by grouping, we can factor the middle term out of the polynomial and then factor the remaining terms separately.

First, the middle term is factored out of the equation. This is done by multiplying the first and last terms together, which in this case is (6t^2)(-4). This results in the equation 6t^2 + 5t - 4 being rewritten as 6t^2 + (5t - 4)(-4).

Next, the remaining terms are factored separately. The first term, 6t^2, is a perfect square and can be factored as (3t)(2t). The second term, (5t - 4)(-4), can be factored by taking out a common factor from each term. In this case, the common factor is (-4). This results in the equation being rewritten as (3t)(2t) + (-4)(5t - 4).

The final step is to group the terms together and factor out the greatest common factor. In this equation, the greatest common factor is (3t)(-4). Thus, the equation 6t^2 + 5t - 4 can be rewritten as (3t)(-4)(2t + 5).

In conclusion, the pair of integers used to rewrite the middle term when factoring 6t^2 + 5t - 4 by grouping is (5t - 4) and (6t^2 + 5t).

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2. Consider a set of vectors of the form (a, 3a, a + b). Determine whether the set is a subspace of R3. If the set is a subspace, give a basis and its dimension. (8 marks total)

Answers

The set of vectors of the form (a, 3a, a + b) is not a subspace of R3.



To be a subspace of R3, a set must satisfy the following conditions:
1. The set must be closed under addition.
2. The set must be closed under scalar multiplication.
3. The set must contain the zero vector.

Let's check if the set satisfies these conditions:

1. Closed under addition:
Let u = (a, 3a, a + b) and v = (c, 3c, c + d) be two vectors in the set. The sum of these vectors is u + v = (a + c, 3a + 3c, a + b + c + d). This vector is not in the set because the third component is not of the form a + b. Therefore, the set is not closed under addition.

2. Closed under scalar multiplication:
Let u = (a, 3a, a + b) be a vector in the set and k be a scalar. The scalar multiplication of u and k is ku = (ka, 3ka, ka + kb). This vector is not in the set because the third component is not of the form a + b. Therefore, the set is not closed under scalar multiplication.

3. Contains the zero vector:
The zero vector in R3 is (0, 0, 0). This vector is not in the set because the third component is not of the form a + b. Therefore, the set does not contain the zero vector.

Since the set does not satisfy the conditions to be a subspace of R3, it is not a subspace. Therefore, there is no basis or dimension for this set.

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If the surface area of a cube is 160 in2, then which best describes the length of an edge of the cube?

Answers

Answer:

5.16 in

Step-by-step explanation:

If the surface area of a cube is 160 in2, then which best describes the length of an edge of the cube?

find the area of ​​one of the 6 squares that form the cube (total area : 6)find the side of the square using the inverse formula L = √A

160 : 6 = 26.66 L = √26.66 = 5.16 in

(2) Solve the inequality |2x − 5| ≤ 9 and present your answer in interval notation.
(3) Find the inverse of the following function:
f(x)= 5 .
6x−1
(4) Letf(x)=√x−4andg(x)=x2−11x+30. Find fg and fg andstatetheirdomains.
(5) Find the equation of the line perpendicular to y = 35 x − 4 and passing through the point (1, 2). Write your answer in slope-intercept form (i.e., y = mx + b).
(6) Divide the following using long division:
2x3 +x2 +3x+5. x−1
Write your final answer in the form Dividend = Quotient + Remainder .

Answers

The final answer is 2x^3 +x^2 +3x+5 = (2x^2 + 3x + 6)(x−1) + 11.

(2) To solve the inequality |2x − 5| ≤ 9, we need to split it into two separate inequalities and solve for x:
2x − 5 ≤ 9 and 2x − 5 ≥ −9
2x ≤ 14 and 2x ≥ 4
x ≤ 7 and x ≥ 2
The solution in interval notation is [2, 7].

(3) To find the inverse of the function f(x) = 5/(6x−1), we need to switch the x and y values and solve for y:
x = 5/(6y−1)
6y−1 = 5/x
6y = 5/x + 1
y = (5/x + 1)/6
The inverse function is f^(-1)(x) = (5/x + 1)/6.

(4) To find fg and fg, we need to plug in the functions for x and simplify:
fg(x) = f(g(x)) = √(x^2−11x+30−4) = √(x^2−11x+26)
fg(x) = g(f(x)) = (√x−4)^2−11(√x−4)+30 = x−8√x+16−11√x+44+30 = x−19√x+90
The domain of fg is all real numbers greater than or equal to 26, and the domain of fg is all real numbers greater than or equal to 0.

(5) To find the equation of the line perpendicular to y = 35 x − 4 and passing through the point (1, 2), we need to find the slope of the perpendicular line and use the point-slope form:
The slope of the original line is 35, so the slope of the perpendicular line is -1/35.
Using the point-slope form, y − y1 = m(x − x1), we get:
y − 2 = −1/35(x − 1)
y = −1/35x + 2 + 1/35
y = −1/35x + 71/35
The equation of the line in slope-intercept form is y = −1/35x + 71/35.

(6) To divide 2x^3 +x^2 +3x+5 by x−1 using long division, we need to divide each term of the dividend by the divisor and find the remainder:
2x^3 ÷ x = 2x^2
2x^2(x−1) = 2x^3−2x^2
(x^2 +3x+5) − (2x^3−2x^2) = 3x^2 +3x+5
3x^2 ÷ x = 3x
3x(x−1) = 3x^2−3x
(3x+5) − (3x^2−3x) = 6x+5
6x ÷ x = 6
6(x−1) = 6x−6
5 − (6x−6) = 11
The final answer is 2x^3 +x^2 +3x+5 = (2x^2 + 3x + 6)(x−1) + 11.

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SOMEONE PLEASE HELP

Answer my question FOR 100 POINTS

Answers

The following solutions with regard to resolving or simplifying the radii of circular ponds are given below.

Part A: 5√100 + √164 meters = 5 × 10 + 2√41 meters

Part B: 5 × 10 + 2√41 meters = 50 + 2√41 meters
Part C: The radius of Pond B is 10√2 meters.
Part D: the radius of Pond B is greater than the radius of Pond A.

What is the explanation for the above?

Part A: The mistake is in Step 1. To simplify the square root of 164, we need to find its prime factors:

164 = 2 × 2 × 41

So, we can write:

√164 = √(2 × 2 × 41) = 2√41

Using this, we can rewrite Step 1 as:

5√100 + √164 meters = 5 × 10 + 2√41 meters

Part B: Using the corrected step from Part A, we can simplify the radius of Pond A as:

5 × 10 + 2√41 meters = 50 + 2√41 meters

So, the radius of Pond A is 50 + 2√41 meters.

Part C: The radius of Pond B is already simplified, so we don't need to do any additional steps. It is:

25√200/5 meters = 5√200 meters

We can simplify this further by finding the prime factorization of 200:

200 = 2 × 2 × 2 × 5 × 5

So, we can write:

√200 = √(2 × 2 × 2 × 5 × 5)

= 2 × 5√2

Using this, we can rewrite the expression for the radius of Pond B as:

5 × 2 × √2 meters

= 10√2 meters

Thus, the radius of Pond B is 10√2 meters.

Part D: We can compare the radii of the ponds using the original expressions by writing:

√164 < 25√200/5

Simplifying both sides:

2√41 < 10√2

Dividing both sides by 2:

√41 < 5√2

Squaring both sides (since both sides are positive):

41 < 25 × 2

41 < 50

So, the inequality is true. Therefore, the radius of Pond B is greater than the radius of Pond A.

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Full Question:

Two circular ponds at a botanical garden have the following radii:

Pond A: Sqrt(164) meters;

Pond B: 25Sqrt(200)/5 meters:

Todd simplified the radius of pond A this way:

5sqrt(164)

Step 1: 5sqrt(100) + Sqrt(164) meters

Step 2: 5(10 + 8)

Step 3: 5(18)

Step 4: 90

One of the steps above is incorrect.

Part A: Rewrite the steps so that it is correct

Part B: Using the corrected step from Part A, simplify the radius of Pond A.

Part C: Simplify the expression for the radius of pond B.

Part D: Write an inequality to compare the radii of the ponds, using the original expressions.

Write a function rule for "The output is three less than the input x ." y=

Answers

Answer:

x to the power of 67000 to the 5th power.

Step-by-step explanation:

Help please i need it by tonight

Answers

Answer:

20

Step-by-step explanation:

The following unit multipliers are used:

1 loaf of bread = 400g of flour = 0.4kg of flour: [tex]\frac{0.4kg Of Flour}{1 loaf}[/tex]

1 day = 64 loaves of bread: [tex]\frac{64 loaves}{1 day}[/tex]

∴Mass of flour needed to last for 20 days:

[tex](20 days)[/tex]× [tex]\frac{64loaves}{1day}[/tex] ×[tex]\frac{0.4 kg}{1 loaf}[/tex]

  = 512 kg of flour

                  1 sack = 25 kg of flour

x sacks of flour = 512 kg of flour

Cross-multiplication is applied:

(x sacks of flour)(25 kg of flour) = (1 sack)(512 kg of flour)

x sacks of flour = [tex]\frac{(1 sack)(512kg Of Flour)}{25 kg Of Flour}[/tex]

                             = 20.48 sacks of flours

∴The minimum number of sacks = 20

use this triangle for problems 4 and 5
1. What is the length of AB
2. What is the area of this triangle​

Answers

(1) The length of AB in triangle ABC 24.58 units.

(2) The area of the triangles is 77.94 sq. units.

What is the length of AB for triangle ABC?

The length of AB in triangle ABC given is calculated by applying cosine rule as shown below.

Length AB is opposite angle C, so will denote the length as c.

c² = a² + b² - 2ab cos(C)

where;

a is the length of side CBb is the length of side ACC is the angle C

c² = (18)² + (10)² - 2(18 x 10) cos(120)

c² = 604

c = √ 604

c = 24.58 units

The area of the triangles is calculated as follows;

A = ¹/₂ab sin C

A = ¹/₂ x 18 x 10 x sin (120)

A = 77.94 sq. units

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Convert the Cartesian coordinates (1,−5) to polar coordinates with r>0 and 0≤θ<2π
r = _______
Enter exact value. θ = _______

Answers

By converting the Cartesian coordinates (1,−5) to polar coordinates with r>0 and 0≤θ<2π r = √26 and θ = 4.909784033953984

To convert the Cartesian coordinates (1,−5) to polar coordinates with r>0 and 0≤θ<2π, we need to use the following formulas:
r = √(x² + y²)
θ = tan⁻¹(y/x)
Plugging in the given values of x = 1 and y = -5, we get:
r = √(1² + (-5)²)
r = √(1 + 25)
r = √26
θ = tan⁻¹(-5/1)
θ = tan⁻¹(-5)
θ = -1.373400766945016
Since we want 0≤θ<2π, we need to add 2π to θ to get the correct value:
θ = -1.373400766945016 + 2π
θ = 4.909784033953984
So the polar coordinates are:
r = √26
θ = 4.909784033953984

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what is the frequency of each interval

Answers

81-100. ------2

1-20----------2

21-40---------4

41-60----------1

61-80----------1

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