If m ABD = 61, and m DBC = 59, then m ABC = ? PLS ASAP

Given: [tex]2x^2 - 12x + 1 = 0[/tex]

Step 1: [tex]2(x^2- 6x + 3^2) + 1 - 2(3^2) = 0[/tex]

Step 2: [tex]2(x^2 - 6x + 9) + 1 - 18 = 0[/tex]

Second answer:  [tex]2(x - 3)^2 - 17 = 0[/tex]

Explanation in detail:

The quadratic equation a[tex]x^{2} + bx + c = 0[/tex]'s vertex form is

where a[tex](x - h)^{2} + k[/tex] equals zero

A is the [tex]x^{2}[/tex] coefficient, and h is the vertex's x-coordinate on the equation's graph.

The vertex of the equation's graph, k, has the y-coordinate.

The completing square can be used to determine the vertex form.

[tex]2x^{2} - 12x + 1 = 0[/tex] is the equation.

Put[tex]2x^{2} -12x[/tex] in a bracket and subtract 2 from them as a common component to utilize the completing square.

∵ [tex]2(x^{2} - 6x) + 1 = 0[/tex]

To determine the product of the first and second terms in the binomial, divide the second term by two.

∵ [tex]6x / 2 = 3x[/tex]

∵[tex]3x = 3 * x[/tex]

∴ The binomial's first and second terms are x and 3, respectively.

The phrase in the bracket's middle is (-)

∴ Middle of the binomial's sign is (-)

∴ The binary value is [tex](x - 3)^2[/tex]

Square 3 is nine.

To keep the equation from altering, you must deduct the same number from the bracket after adding 9 there.

∴ [tex]2(x^2 - 6x + 9 - 9) + 1 = 0[/tex]

- Remove the brace and multiply -9 by 2.

∵ [tex]2 *-9 = -18[/tex]

∴ [tex]2(x^2- 6x + 9) + 1 - 18 = 0[/tex]

- Construct a square binomial from [tex](x^{2} - 6x + 9)[/tex]

∵ [tex]x^2 - 6x + 9 = (x - 3)^2[/tex]

∴ [tex]2(x - 3)^2 + 1 - 18 = 0[/tex]

Include related terms.

∴[tex]2(x - 3)^2 - 17 = 0[/tex]

∴ The equation's vertex form is[tex]2(x - 3)^2 - 17 = 0.[/tex]

Given that: [tex]2x^2 - 12x + 1 = 0[/tex]

Step 1: [tex]2(x^2- 6x + 3^2) + 1 - 2(3^2) = 0[/tex]

Step 2: [tex]2(x^2 - 6x + 9) + 1 - 18 = 0[/tex]

Step 3:  [tex]2(x - 3)^2 - 17 = 0[/tex]

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Each side of equilateral triangle will have the length equals to 42.7 cm.

There are different types of quadrilaterals, for example, square, rectangle, rhombus, trapezoid, and parallelogram.  Each type is defined accordingly to its length of sides and angles. For example, in a rectangle,  the opposite sides are equal and parallel and their interior angles are equal to 90°.

The perimeter of a geometric figure is the sum of its sides. Thus, for a rectangle, the perimeter is 2L+2W.

You should find the perimeter of the rectangle.

2P=23+23+41+41= 128 cm

The equilateral triangle will have the same perimeter of the rectangle. An equilateral triangle presents the equal three sides, thus,

128= 3L

L=128/3=42.7 cm

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The solution to the system and the solution that fills the blank space is (6, 1)

This are also known as a set of simultaneous equations, also known as a system of equations which is a finite set of equations for which common solutions

Given the following system of equations

2y − x = 8, and

y − 2x = −5.

From equation 2;

y = -5 + 2x ............. 3

Substitute equation 2 into equation 1 to have:

2(-5+2x) - x = 8

Expand to have:

-10 +4x -x = 8

-10 + 3x = 8

3x = 8 + 10

3x = 18

Divide both sides

by 3

3x/3 = 18/3

x = 6

Substitute x = 6 into equation 3

y = -5 + 2x

y = -5 + 2(3)

y = -5 + 6

y = 1

Hence the solution to the system and the solution that fills the blank space is (6, 1)

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Answer:

Its D (6,7)

Step-by-step explanation:

The question requires that we have to state the hypothesis:

null hypothesis;H0: u1 < u2, the alternate hypothesis;H1: u1 > u2. The one tail test is what is to be used.

H0: u1 < u2

H1: u1 > u2

The one tail test statistic is what is to be used here

[tex]\sqrt{\frac{30.8}{17} +\frac{6.1}{15} }[/tex]

= 1.49

The df = 17 + 15 - 2

= 30

test statistic = 18.9 - 17.4 / 1.49

= 1.007

We have the critical value on excel as T.INV(0.9,30)

=  1.310

E. At 0.1, we can conclude that  t-value (1.007) does not lies in the rejection area. We fail to reject the null hypothesis. Hence we conclude that the mean vacation with the unlimited plan is greater.

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Answer:

y^2 = -8x  for example.

Step-by-step explanation:

General form is y^2 = -4ax

The focus of this parabola will be at (-a , 0) on the x axis.

So the parabola might be:

y^2 = -4x

y^2 = -12x

y^2 = -0.9x  and so on......

A semicircle is a part of a circle, and it is referred to as half of a given circle. Thud the area of the semi-circle window is 982 [tex]in^{2}[/tex].

A circle is a shape that is bounded by a curved path which is referred to as the circumference. Some parts of a circle are radius, diameter, sector, arc, semi-circle, circumference, etc.

A semicircle is a part of a circle, and it is referred to as half of a given circle.

such that:

Area of a circle = [tex]\pi[/tex] [tex]r^{2}[/tex]

and

area of a semicircle = [tex]\frac{\pi r^{2} }{2}[/tex]

where: r is the radius of the circle, and [tex]\pi[/tex] is a constant with a value of [tex]\frac{22}{7}[/tex].

Thus from the given question, it can be inferred that;

r = [tex]\frac{50}{2}[/tex]

 = 25

r = 25 in

Thus, the area of the semi-circle can be determined as;

area of the semi-circle = [tex]\frac{1}{2}[/tex] * [tex]\frac{22}{7}[/tex] * [tex]25^{2}[/tex]

                                      = 982.1429

area of the semi-circle = 982.14 [tex]in^{2}[/tex]

The area of the semi-circle window is approximately 982 [tex]in^{2}[/tex].

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The answer is A. Winter.

As the pattern is : F, W, Sp, Su, F... (letters used for convenience), then the next word in the pattern will be Winter, because that was shown in the previous sequence.

dis i know functiin of fx is ♾

Answer:

4th option

Step-by-step explanation:

the end behaviour is what happens when x gets larger and positive (right- hand end )or larger and negative ( left- hand end )

f(x) = [tex]\frac{1}{2}[/tex] x² ← is of even degree (2) and has a positive leading coefficient , then

• even degree and positive leading coefficient

f(x) → +∞ as x → + ∞ and

f(x) → + ∞ as x → - ∞

The complete proof from the figure is:

The given parameter is:

AC ≅ DF; AB ≅ DE; ∠1 ≅ ∠6

Angles 1 and 2, & Angles 5 and 6 are vertical angles

So, we have: ∠1 ≅ ∠2; ∠5 ≅ ∠6 by the definition of vertical angles.

From the figure, triangles ABC and DEF are congruent by the SAS postulate.

So, we have: ΔABC ≅ ΔDEF by the definition of SAS postulate

Angles 3 and 4 are corresponding angles

So, we have: ∠1 ≅ ∠2; ∠5 ≅ ∠6 by the definition of corresponding angles.

Hence, the lines BC and EF are parallel lines

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Answer:

the number is 2

Step-by-step explanation:

5x = x + 8

5x - x = 8

4x = 8

x = 8/4

x = 2

Answer:

109x

Step-by-step explanation:

We know that there are 109 students in the fourth grade, who each have x cupcakes. In order to find the total amount of cupcakes, we could add all of the cupcakes together to get:

x + x + x + x + .... x

109 times because there are 109 students.

This can be simplified to become 109x.

Answer:

Step-by-step explanation:

The most basic quadratic function is f(x) = x2, whose graph appears below. ... us to graph an entire family of quadratic functions using transformations.

Answer:

50 ^^

Step-by-step explanation:

So first an easy way to solve this problem is by removing the zeros then you'll have 35 and
7x1=7

7x2=14

7x3=21

7x4=28

7x5=35

So 7x5 dont forget to add your zeros to 35 and 5 and boom your answer is 50

need more explaining tell me ^^

Answer:

A.  y = 1/3x^2

Step-by-step explanation:

The red graph is stretched horizontally by a factor 1/3  to give the blue graph

The first quartile of the dataset {1, 2, 3, 4, 5, 6} is 2.

Given data set: {1,2,3,4,5,6}

Median of the given data set =3.5

The data set can be split into 2 parts as, {1,2,3} and {4,5,6}. The first quartile is calculated by simply finding the median of the first part of the data set.

Thus, the first quartile is 2.

The quarter divides the distribution into four groups and calculates the range of values above and below the mean.

A quartile separates the dataset into four categories by dividing the data into three points: the lowest, median, and upper quartiles.

The interquartile range, a measurement of variation around the median, is calculated using quartiles.

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Answer:

[tex]\textsf{A. } y=3x^2+10x-8[/tex]

Step-by-step explanation:

We are given the information that a function has zeros at x = ⅔ and x = -4. In order to find the function that has those zeros, we can substitute the value of the zeros into each function. If the value of a function equates to zero with both values, then that is the function we are looking for.

..................................................................................................................................................

Standard Form of a Quadratic: ax² + bx + c = 0.

..................................................................................................................................................

[tex]\large \text{$y = 3x^2+10x-8 \implies 0=3x^2+10x-8$}[/tex]

[tex]\boxed{\begin{minipage}{15 em}{\text{$x=\dfrac{2}{3}$}} \\ \\\implies 0=3\left(\dfrac{2}{3}\right)^2+10\left(\dfrac{2}{3}\right)-8\\\\\implies 0=3\left(\dfrac{4}{9}\right)+10\left(\dfrac{2}{3}\right)-8\\\\\implies 0=\not{3}\left(\dfrac{4}{\not{9}\ 3}\right)+\dfrac{20}{3}-8\\\\\implies 0=\dfrac{4}{3}+\dfrac{20}{3}-8\\\\\implies 0=\dfrac{24}{3}-8\\\\\implies 0=0\ \checkmark\end{minipage}}[/tex]    [tex]\boxed{\begin{minipage}{15 em}{\text{$x=-4$}} \\ \\\implies 0=3(-4)^2+10(-4)-8\\\\\implies 0=3(16)-40-8\\\\\implies 0=48-48\\\\\implies 0=0\ \checkmark\end{minipage}}[/tex]

..................................................................................................................................................

[tex]\large \text{$y = 2x^2-5x-12 \implies 0=2x^2-5x-12$}[/tex]

[tex]\boxed{\begin{minipage}{15 em}{\text{$x=\dfrac{2}{3}$}} \\ \\\implies 0=2\left(\dfrac{2}{3}\right)^2-5\left(\dfrac{2}{3}\right)-12\\\\\implies 0=2\left(\dfrac{4}{9}\right)-\dfrac{10}{3}-12\\\\\implies 0=\dfrac{8}{9}-\dfrac{10}{3}-12\\\\\implies 0=\dfrac{8}{9}-\dfrac{10\times3}{3\times3}-\dfrac{12\times9}{9}\\\\\implies 0=\dfrac{8}{9}-\dfrac{30}{9}-\dfrac{108}{9}\\\\\implies0=-\dfrac{22}{9}-\dfrac{108}{9}\\\\\implies 0=-\dfrac{130}{9}\ \textsf{X}\end {minipage}}[/tex]    [tex]\boxed{\begin{minipage}{15 em}{\text{$x=-4$}} \\ \\\implies 0=2(-4)^2-5(-4)-12\\\\\implies 0=2(16)+20-12\\\\\implies 0=32+20-12\\\\\implies 0=40\ \textsf{X}\end{minipage}}[/tex]

..................................................................................................................................................

[tex]\large \text{$y = 2x^2+5x-12 \implies 0=2x^2+5x-12$}[/tex]

[tex]\boxed{\begin{minipage}{15 em}{\text{$x=\dfrac{2}{3}$}} \\ \\\implies 0=2\left(\dfrac{2}{3}\right)^2+5\left(\dfrac{2}{3}\right)-12\\\\\implies 0=\dfrac{8}{9}\right)+\dfrac{10}{3}-12\\\\\implies 0=\dfrac{8}{9}\right)+\dfrac{10\times3}{3\times3}-\dfrac{12\times9}{9}\\\\\implies 0=\dfrac{8}{9}+\dfrac{30}{9}-\dfrac{108}{9}\\\\\implies0=\dfrac{38}{9}-\dfrac{108}{9}\\\\\implies 0=-\dfrac{70}{9}\ \textsf{X}\end{minipage}}[/tex]    [tex]\boxed{\begin{minipage}{15 em}{\text{$x=-4$}} \\ \\\implies 0=2(-4)^2+5(-4)-12\\\\\implies 0=2(16)-20-12\\\\\implies 0=32-32\\\\\implies 0=0\ \checkmark\end{minipage}}[/tex]

..................................................................................................................................................

[tex]\large \text{$y = 3x^2-10x-8 \implies 0=3x^2-10x-8$}[/tex]

[tex]\boxed{\begin{minipage}{15 em}{\text{$x=\dfrac{2}{3}$}} \\ \\\implies 0=3\left(\dfrac{2}{3}\right)^2-10\left(\dfrac{2}{3}\right)-8\\\\\implies 0=3\left(\dfrac{4}{9}\right)-\dfrac{20}{3}-8\\\\\implies 0=\not{3}\left(\dfrac{4}{\not{9}\ 3}\right)-\dfrac{20}{3}-8\\\\\implies 0=\dfrac{4}{3}-\dfrac{20}{3}-\dfrac{8\times3}{3}\\\\\implies 0=-\dfrac{16}{3}-\dfrac{24}{3}\\\\\implies 0=-\dfrac{40}{3}\ \textsf{X}\end{minipage}}[/tex]    [tex]\boxed{\begin{minipage}{15 em}{\text{$x=-4$}} \\ \\\implies 0=3(-4)^2-10(-4)-8\\\\\implies 0=3(16)+40-8\\\\\implies 0=48+40-8\\\\\implies 0=80\ \textsf{X}\end{minipage}}[/tex]

..................................................................................................................................................

Therefore, the function whose zeros are ⅔ and -4 is [tex]y=3x^2+10x-8[/tex].

By substitution, the value of x and y are 1 and 1 respectively.

3 / 8 x + 1 /3  y = 17 / 24

x + 7y = 8

x = 8 - 7y

Let's substitute the value of x in equation(ii)

3 / 8(8 - 7y) + 1 / 3 y = 17 / 24

3 - 21 / 8 y + 1 / 3 y = 17 / 24

1 / 3 y - 21 / 8 y = 17 / 24 - 3

8y - 63y/24 =  17 - 72/ 24

-55y / 24 = -55 / 24

cross multiply

-1320y = -1320

divide both sides by -1320

y = -1320 / -1320

y = 1

Hence,

x + 7(1) = 8

x = 8 - 7

x = 1

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Answer:81.85 km/hr is the average speed

Step-by-step explanation:

Answer: ≈ 84,3 km/h.

Step-by-step explanation:

[tex]\displaystyle\\V{average}=\frac{\Delta S}{\Delta t}=\frac{S_1+S_2+S_3}{t_1+t_1+t_3} .\\1.\ S_1=50\ km\\ t_1=\frac{S_1}{V_1}\\ V_1=25*\frac{3600}{1000} \\V_1=90\ \frac{km}{h} .\\t_1=\frac{50}{90}\\ t_1=\frac{5}{9} \ h.\\2.\ S_2=120\ km\ \ \ \ V_2=80\ \frac{km}{h} \\t_2=\frac{S_2}{V_2} \\t_2=\frac{120}{80} \\t_2=\frac{3}{2}\ h.\\[/tex]

[tex]3.\ V_3=90\ \frac{km}{h} \\t_3=35\ minutes\\t_3=\frac{35}{60} \\t_3=\frac{7}{12}\ h.\\ S_3=V_3*t_3\\S_3=90*\frac{7}{12} \\S_3=52,5\ km.\\Hence,\\V{average}=\frac{50+120+52,5}{\frac{5}{9}+\frac{3}{2} +\frac{7}{12} } \\V{average}=\frac{222,5}{\frac{95}{36} } \\Vaverege\approx84,3\ \frac{km}{h} .[/tex]

The first three elements of the recursive series are 6, 10, 8. (Correct choice: B)

In this question we have a kind of recursive function known as Fibonacci's function, where a value of the series is generated from at least immediately previous elements. In this case, we need to find the first three elements from the fifth and fourth elements of the series:

[tex]a_{n-2} = a_{n-1} - a_{n} + 4[/tex]

a₄ = a₅ - a₆ + 4

a₄ = - 2 - 0 + 4

a₄ = 2

a₃ = a₄ - a₅ + 4

a₃ = 2 - (- 2) + 4

a₃ = 8

a₂ = a₃ - a₄ + 4

a₂ = 8 - 2 + 4

a₂ = 10

a₁ = a₂ - a₃ + 4

a₁ = 10 - 8 + 4

a₁ = 6

The first three elements of the recursive series are 6, 10, 8. (Correct choice: B)

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Answer:

The second choice is correct.

Step-by-step explanation:

As we go right on the number line, we are getting to larger and larger values.

D is greater than C because it is to the right of C, so the first choice is not correct.

The third choice is showing absolute value.  This is saying that the letters C and D are both the same distance from zero. They are not.

The last choice says that A is to the right of B and it is not.

The second choice that is B≥ C is correct.

The basics of this question requires us to read the number line perfectly

As we go right on the number line, we are getting to larger and larger values.

D is greater than C because it is to the right of C, so the first choice is not correct.

The third choice is showing absolute value.  This is saying that the letters C and D are both the same distance from zero. They are not.

The last choice says that A is to the right of B and it is not.

Thus option 2) is correct

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Answer:

Step-by-step explanation:

Begin by changing the radical form to exponential form:

[tex]((1024w^{-2}x^{-3}y^6z^{-1)^2)^{\frac{1}{5} }[/tex]  There are a couple of ways to do this, but let's distribute the power of 2 into the parenthesis first by using the power to power rule of multiplication:

[tex](1024^2w^{-4}x^{-6}y^{12}z^{-2})^{\frac{1}{5}[/tex]  Now we can use the same rule of power to power and distribute the power of 1/5 in:

[tex]1024^{\frac{2}{5}}w^{-\frac{4}{5}}x^{-\frac{6}{5}}y^{\frac{12}{5}}z^{-\frac{2}{5}[/tex]

The first term simplifies down to 16, and then in order to make the negative exponents positive, we move them to the denominator of a fraction, leaving the positive one where it is:

[tex]\frac{16y^{\frac{12}{5}} }{w^{\frac{4}{5}}x^{\frac{6}{5} }z^{\frac{2}{5} } }[/tex] which is the third choice down

Answer:

LK = 4[tex]\sqrt{2}[/tex]

Step-by-step explanation:

given point J (4, 4 )

then LJ = 4 and JK = 4

using Pythagoras' identity in right triangle LJK

LK² = LJ² + JK² = 4² + 4² = 16 + 16 = 32 ( take square root of both sides )

LK = [tex]\sqrt{32}[/tex] = [tex]\sqrt{16(2)}[/tex] = [tex]\sqrt{16}[/tex] × [tex]\sqrt{2}[/tex] = 4[tex]\sqrt{2}[/tex]

The area of the given semicircle is 56.574 sq. units. Half of the area of the circle is the area of the semicircle.

The area of the semicircle is

A = [tex]\frac{1}{2}[/tex] πr² sq. units

Where r - radius

A semicircle is nothing but half part of the circle. So, its area is also half of the circle's area.

It is given that,

The radius of the semicircle is

r = 6

So, its area in terms of π is

A = [tex]\frac{1}{2}[/tex] πr²

   =  [tex]\frac{1}{2}[/tex] × π × 6²

   = 18π sq. units

On substituting π = 3.143, we get

A = 18 × 3.143 = 56.574 sq. units

Thus, the area of the given semicircle in terms of π is 18π or in terms of π =3.143 is 56.574 sq. units.

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The value of x in the system of equation is 6

The system of equations is given as:

y = 1/2 * 2^x

y = 5x + 2

Substitute y = 1/2 * 2^x in y = 5x + 2

1/2 * 2^x = 5x + 2

Multiply through by 2

2^x = 10x + 4

Next, we use the trial by error.

Set x = 6

This gives

2^6 = 10 * 6 + 4

Evaluate the product and the exponent

64 = 60 + 4

Evaluate the sum

64 = 64

Hence, the value of x in the system of equation is 6

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The addition equation to represent how their charges combine to make the overall charge of the atom is + 1 + - 1 = 0

A hydrogen atom has one positively charged proton and one negatively charged electron.

An atom of every elements has an electron(negatively charged) and positively charge(proton) part.

An atom as a whole is electrically neutral if the number of proton and electron are equal. They will cancel each other to make the atom neutral.

Therefore, the addition equation to represent how their charges combine to make the overall charge of the atom is as follows;

+ 1 + - 1 = 0

1 positively charged proton plus single negatively charged electron  = zero(neutral)

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Answer: 2

Step-by-step explanation:

remember y=mx^2+c

where m = gradient = slope

∴ expand and simplify y-5=2(x+3)

y-5=2x+6

y=2x+11

m=2  ∴ slope: 2

Answer:

D. 13

Step-by-step explanation:

54 ÷ 3 = 18 (your age in 5 years)

18 - 5 = 13 (your age now)

Answer:

Step-by-step explanation:

The answer is Economists

Answer: economists

Step-by-step explanation: got it right

In the given right triangle, the trigonometric ratio, sin J has a fractional value of 5/13 and a decimal value of 0.39.

In trigonometry, for a right triangle, the sine (sin) of any angle θ is given as the ratio of its opposite side to the hypotenuse of the triangle, that is, sin θ = (opposite side)/(hypotenuse).

In the question, we are asked to find the trigonometric ratio, sin J, for the given right triangle JKL.

The side opposite to angle J is KL, which has a value of 3 units.

The hypotenuse of the given right triangle is JK, which has a value of 7.8 units.

Thus, sin J can be calculated as:

sin J = KL/JK = 3/7.8 = 5/13 = 0.39.

Thus, in the given right triangle, the trigonometric ratio, sin J has a fractional value of 5/13 and a decimal value of 0.39.

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Answer:

E. >

Because 65 is greater than 56.