√3018+√36+√169 can you solve this for me please

Answer:

C

Step-by-step explanation:

This is a translation 5 units left and 5 units dowh.

Here, the translation is [tex]T < 5,-5 >[/tex].

Here, the translation is given as: [tex]y=2(x-5)+5\longrightarrow y=2(x-0)+0[/tex].

i.e. [tex]y=2(x-5)+5\longrightarrow y=2(x-5+5)+(5-5)[/tex].

So, the graph of [tex]y[/tex] moves 5 units to the right and (-5) units to the up i.e., 5 units to the down.

Therefore, here, the translation is [tex]T < 5,-5 >[/tex].

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In mathematics a linear inequality is an inequality involving a linear function. A linear inequality contains one of the inequality symbols.​< is less than > is greater than ≤ is less than or equal to ≥ is greater than or equal to ≠ is not equal to.

Let's find the critical points of the inequality.

x² − 9x −(−8) = − 8 −(−8) (Subtract -8 from both sides)

(x − 1)(x − 8) = 0 (Factor left side of equation)

x − 1 = 0 or x − 8 = 0 (Set factors equal to 0)

x < 1 (Doesn't work in original inequality)

1 < x < 8 (Works in original inequality)

x > 8 (Doesn't work in original inequality)

The third option is the correct one.

The value of ∠B is 60 degrees

The given parameters are

2∠A = 3∠B = 6∠C

Let the angles in a triangle ABC be ∠A, ∠B and ∠C.

The sum of these angles in the triangle ABC is

∠A + ∠B + ∠C = 180

Multiply the above equation by 2

2 * (∠A + ∠B + ∠C) = 2 * 180

Evaluate the product

2∠A + 2∠B + 2∠C = 360

Recall that 2∠A = 3∠B.

Substitute the above in the equation 2∠A + 2∠B + 2∠C = 360

3∠B + 2∠B + 2∠C = 360

This gives

5∠B + 2∠C = 360

Multiply the above equation by 3

3 * (5∠B + 2∠C) = 3 * 360

Evaluate the product

15∠B + 6∠C = 1080

Recall that 3∠B = 6∠C.

Substitute the above in the equation

So, we have:

15∠B + 3∠B = 1080

Evaluate the like terms

18∠B = 1080

Divide both sides by 18

∠B = 60

Hence, the value of ∠B is 60 degrees

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The equation of the line passing through the point (-5, -7) with a slope of 2/5 is y = (2/5)x - 5.

To find the equation of a line, we can use the point-slope form of a linear equation:

y - y₁ = m(x - x₁),

where (x₁, y₁) is the given point on the line and m is the slope.

In this case, the given point is (-5, -7) and the slope is 2/5. Substituting these values into the equation, we have:

y - (-7) = (2/5)(x - (-5)).

Simplifying further:

y + 7 = (2/5)(x + 5).

Distributing the 2/5:

y + 7 = (2/5)x + 2.

Subtracting 7 from both sides:

y = (2/5)x - 5.

Therefore, the equation of the line passing through the point (-5, -7) with a slope of 2/5 is y = (2/5)x - 5.

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[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]

For two lines to be parallel, there should be angles that follow some specific properties that is usually observed with parallel lines.

We can clearly see that :

[tex] \qquad❖ \: \sf \: \angle7 \cong \angle16[/tex]

( by Alternate interior angle pair )

[tex] \qquad \large \sf {Conclusion} : [/tex]

Lines l and m are parallel to each other.

Answer:

The milk would be 60 ml and the topping would be 30 ml

Step-by-step explanation:

If there are 2 parts milk and 1 part toppings that would be a total of 3 (2+1 =3)  So we are looking for 2/3 of 90 and 1/3 of 90.

Using Hinton's method;

A four-dimensional shape (4D) is a mathematical extension of a three-dimensional or 3D space.

Three-dimensional space is the simplest possible abstraction of the observation that one only needs three numbers, called dimensions, to describe the sizes or locations of objects.

Using Hinton's method;

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The answer is -3.

If m = 4 and n = -7, we can evaluate the expression by substitution.

m + n

4 + (-7)

-3

❄ H there,

evaluate this expression by substituting the provided parameters –

[tex]\sf{m+n} \ | \ m=4 \ \& \ n=-7 \ | 4+(-7)=4-7=-3}[/tex]

That's it!

❄

A statement correctly compares functions f and g is that: C. they have the same end behavior as x approaches -∞ but different end behavior as x approaches ∞.

A function can be defined as a mathematical expression that defines and represents the relationship between two or more variable, which is typically modelled as input (x-values) and output (y-values).

In Mathematics, there are different types of functions and these include the following;

Function g is represented by the following table and a line representing these data is plotted in the graph that is shown in the image attached below.

x          -1   0   1    2   3   4

g(x)     24  6   0  -2  

Based on the line, we can logically deduce the following points:

This ultimately implies that, a statement correctly compares functions f and g is that both functions have the same end behavior as x approaches -∞ but different end behavior as x approaches ∞.

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Using the relation between velocity, distance and time, it is found that the rate of the current is of 3.33 mph.

Velocity is distance divided by time, hence:

v = d/t

A small motorboat travels 12mph in still water. With the current, upstream, 46 miles are traveled in t hours, hence:

12 + r = 46/t

r = 46/t - 12

Downstream, the time is of t + 2 hours, hence:

12 - r = 46/(t + 2)

r = 12 - 46/(t + 2)

Hence, equaling the values for r:

46/t - 12 = 12 - 46/(t + 2)

46/t + 46/(t + 2) = 24

[tex]\frac{46t + 92 + 46t}{t(t + 2)} = 24[/tex]

92t + 92 = 24t² + 48t

24t² - 44t - 92 = 0

Using a quadratic equation calculator, the solution is t = 3. Hence the rate is found as follows:

r = 46/t - 12 = 46/3 - 12 = 3.33 mph.

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Answer:

an angle bisector

Step-by-step explanation:

This is showing the construction of the bisector of ∠LNM

The function notation form of the given equation; y = -3x +3 as in the task content is; f(x) = -3(x) +3.

According to the task content, the equation given; y = -3x +3 is to be written in function notation.

Consequently, since the function expression given is linear in variable X, it follows that the required function notation is;

f(x) = -3x +3.

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Answer:

x=18

Step-by-step explanation:

[tex]\frac{4}{12} =\frac{6}{x} \\[/tex]

[tex]\frac{4}{12}[/tex] ÷ [tex]\frac{2}{2}[/tex] = [tex]\frac{2}{6}[/tex] = [tex]\frac{6}{x}[/tex]

[tex]\frac{2}{6} = \frac{6}{x}[/tex]  

2×3 =6

6×3=18

x=18

4. Compute the derivative.

[tex]y = 2x^2 - x - 1 \implies \dfrac{dy}{dx} = 4x - 1[/tex]

Find when the gradient is 7.

[tex]4x - 1 = 7 \implies 4x = 8 \implies x = 2[/tex]

Evaluate [tex]y[/tex] at this point.

[tex]y = 2\cdot2^2-2-1 = 5[/tex]

The point we want is then (2, 5).

5. The curve crosses the [tex]x[/tex]-axis when [tex]y=0[/tex]. We have

[tex]y = \dfrac{x - 4}x = 1 - \dfrac4x = 0 \implies \dfrac4x = 1 \implies x = 4[/tex]

Compute the derivative.

[tex]y = 1 - \dfrac4x \implies \dfrac{dy}{dx} = -\dfrac4{x^2}[/tex]

At the point we want, the gradient is

[tex]\dfrac{dy}{dx}\bigg|_{x=4} = -\dfrac4{4^2} = \boxed{-\dfrac14}[/tex]

6. The curve crosses the [tex]y[/tex]-axis when [tex]x=0[/tex]. Compute the derivative.

[tex]\dfrac{dy}{dx} = 3x^2 - 4x + 5[/tex]

When [tex]x=0[/tex], the gradient is

[tex]\dfrac{dy}{dx}\bigg|_{x=0} = 3\cdot0^2 - 4\cdot0 + 5 = \boxed{5}[/tex]

7. Set [tex]y=5[/tex] and solve for [tex]x[/tex]. The curve and line meet when

[tex]5 = 2x^2 + 7x - 4 \implies 2x^2 + 7x - 9 = (x - 1)(2x+9) = 0 \implies x=1 \text{ or } x = -\dfrac92[/tex]

Compute the derivative (for the curve) and evaluate it at these [tex]x[/tex] values.

[tex]\dfrac{dy}{dx} = 4x + 7[/tex]

[tex]\dfrac{dy}{dx}\bigg|_{x=1} = 4\cdot1+7 = \boxed{11}[/tex]

[tex]\dfrac{dy}{dx}\bigg|_{x=-9/2} = 4\cdot\left(-\dfrac92\right)+7=\boxed{-11}[/tex]

8. Compute the derivative.

[tex]y = ax^2 + bx \implies \dfrac{dy}{dx} = 2ax + b[/tex]

The gradient is 8 when [tex]x=2[/tex], so

[tex]2a\cdot2 + b = 8 \implies 4a + b = 8[/tex]

and the gradient is -10 when [tex]x=-1[/tex], so

[tex]2a\cdot(-1) + b = -10 \implies -2a + b = -10[/tex]

Solve for [tex]a[/tex] and [tex]b[/tex]. Eliminating [tex]b[/tex], we have

[tex](4a + b) - (-2a + b) = 8 - (-10) \implies 6a = 18 \implies \boxed{a=3}[/tex]

so that

[tex]4\cdot3+b = 8 \implies 12 + b = 8 \implies \boxed{b = -4}[/tex].

The required probabilities are:

A) P(D or 5) = 4/13

B) P(H and J) = 1/13

C) P(J or 8) = 2/13

D) P(H or S) = 1/2

E) P(R and F) = 3/26

F) P(R or Q) = 7/13

The ratio of favorable outcomes to the total outcomes of an event is said to be its probability.

P(E) = n(E)/n(S)

It is given that a single card is drawn at random from a standard deck of 52 cards.

So, the sample space consists of 52 cards in total

From those,

4 suits: Hearts, Clubs, Spades, Diamonds

Each of the suit has 13 cards: { Ace, 2,3,4,5,6,7,8,9,10, Jack, Queen, King}

There are 26 Red cards and 26 Black cards.

A) The probability of drawing a diamond or a 5:

P(D or 5) = P(D) + P(5) - P(D and 5)

               = 13/52 + 4/52 - 1/52

               = 16/52 = 4/13

B) The probability of drawing a heart and a jack:

P(H and J) = P(H) × P(J) (Since they are independent events)

                 = 13/52 × 4/13

                 = 1/13

C) The probability of drawing a jack or 8:

P(J or 8) = P(J) + P(8) - P(J and 8)

              = 4/52 + 4/52 - 0

              = 2/13

D) The probability of drawing a heart or a spade:

P(H or S) = P(H) + P(S) - P(H and S)

               = 13/52 + 13/52 - 0

               = 26/52 = 1/2

E) The probability of drawing a red and face card:

P(R and F) = P(R) × P(F) (Since they are independent)

                 = 26/52 × 12/52

                 = 1/2 × 3/13

                 = 3/26

(There are three face cards- jack, king, and queen: each of 4)

F) The probability of drawing a red card or a queen:

P(R or Q) = P(R) + P(Q) -P(R and Q)

               = 26/52 + 4/52 - 2/52

               = 28/52 = 7/13

Thus, the required probabilities are calculated.

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The responses include:

In mathematics, a real number simply means a value of a continuous quantity which can represent a distance along a line.

In this case, real numbers are the numbers that include both rational and irrational numbers. Here, Rmrational numbers such as integers (-2, 0, 1), fractions(1/2, 2.5) and the irrational numbers such as √3, π(22/7), etc., are all real numbers.

In conclusion, the correct options are A, B, and C.

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The equation become u = k p^2 / d.

According to the statement

we have given that some conditions for the equations and we have t make the equation from the given equations.

So, For this purpose, we have given that

The equation in which u is varies directly with the square of p and inversely with d and use the k as the constant of proportionality.

So, From all these above the equation will become is

u = k p^2 / d

In latex form the equation write as :

[tex]u = \frac{kp^{2}}{d}[/tex]

In this equation all the given conditions are applicable.

So, The equation become u = k p^2 / d

Disclaimer: This question was incomplete. Please find the full content below.

Question:

Write an equation that expresses the following relationship. u varies directly with the square of p and inversely with d In your equation, use k as the constant of proportionality.

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(8) The algebraic expression representing the given phrase is 22x ≥ 350, where x is the number of units sold by Ms. Reed.

(9) The minimum number of shares needed to achieve the required profit is 1223, using the algebraic expression 2.25x ≥ 2750, where x is the number of shares.

(10) The number of books needed to be sold for the novelist to make a profit of $10,000 is 4350, using the algebraic expression 5000 + 1.1495x ≥ 10000, where x is the number of books sold,

(8) Weekly target for Ms. Reed is $350.

The cost of each unit she sells is $22.

We assume the number of units she sold to be x.

Thus, the total sales done by Ms. Reed is $22x.

For her to remain employed, her total sales should exceed her target, which can be shown as an algebraic expression: 22x ≥ 350.

Thus, the algebraic expression representing the given phrase is 22x ≥ 350, where x is the number of units sold by Ms. Reed.

(9) Profit on each share is $2.

The additional profit is $0.25.

Thus, the total profit on each share is $2 + $0.25 = $2.25.

The required profit by the customer is $2750.

We assume the number of shares needed to be x.

Thus, the total profit made by the customer is $2.25x.

For the customer to make the required profit, we can write the algebraic expression, 2.25x ≥ 2750.

To solve this, we divide both sides by 2.25 to get:

2.25x/2.25 ≥ 2750/2.25,

or, x ≥ 1222.22.

Thus, the minimum number of shares needed to achieve the required profit is 1223.

(10) Default contract of Book Maker publisher is $5000 and 5% royalty.

The cost of the book, for which the novelist has got the contract is $22.99.

We assume the number of books sold to be x.

The royalty share on each book given to the novelist is 5% of $22.99, or, $ 5/100 * 2299/100 = $ 11495/10000 = $1.1495.

Thus, the royalty received on x number of books = $1.1495*x = $1.1495x.

Thus, the total profit to the novelist = (5000 + 1.1495x).

Since the novelist wants to make a minimum profit of $10000, we can show it as the algebraic expression:

5000 + 1.1495x ≥ 10000.

To solve this, we go as follows:

5000 + 1.1495 ≥ 10000,

or, 1.1495x ≥ 10000 - 5000,

or, 1.1495x ≥ 5000,

or, x ≥ 5000/1.1495,

or, x ≥ 4349.717.

Approximating, we get x ≥ 4350.

Thus, 4350 books need to be sold to achieve the wanted profit.

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The function that best fits the graph points is; B: y = 65.0778 * 772.9605ˣ

From the given graph, we can see that is parabolic form and as such we can say it is an exponential function.

Looking at the options, only option B is in exponential form and as such, we will take one point on the graph to check if this is the right function.

Let us use the coordinate (0.9, 26000)

y = 65.0778 * 772.9605ˣ

y = 65.0778 * 772.9605^(0.9)

y = 25837.76

This is very close and as such is the correct option.

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Answer:

  1.6 times   or   60% more than

Step-by-step explanation:

The question seems to be asking about the growth factor in the given exponential function.

A generic exponential function will have the form ...

  quantity = (initial value) × (growth factor)^(number of intervals)

Comparing this form to the given formula ...

  f(x) = 58 × 1.6^x

we see the "growth factor" is 1.6. This is the multiplier from one interval (year) to the next.

Each year the expected number of birds is 1.6 times the number the year before.

__

Additional comment

A growth factor is sometimes expressed in terms of a growth rate, usually a percentage.

  growth factor = 1 + growth rate

  1.6 = 1 + 0.60 = 1 + 60%

The growth rate of this bird population is 60% per year. Each year, the population is 60% more than the year before.

Answer: 1.6

Step-by-step explanation:

The possible formula for the polynomial in discuss whose roots are described as; having roots of multiplicity 2 at x = 3 and x = 0 , and a root of multiplicity 1 at x = − 1 is; P(x) = x^5 -5x⁴-6x³+18x².

The polynomial which is as described in the task content whose roots are as given can be written in its factorised form as follows;

P(x) = (x-3) (x-3) (x) (x) (x+1)

The expanded form is therefore;

P(x) = x^5 - 5x⁴- 6x³+ 18x².

Therefore, the polynomial having roots of multiplicity 2 at x = 3 and x = 0 , and a root of multiplicity 1 at x = − 1 is P(x) = x^5 - 5x⁴- 6x³+ 18x².

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Answer:

f(y-2)+f(4-y)=4

Step-by-step explanation:

Assume (let) x=y-2

So: y=x+2

f(y-2)+f(4-y)=f(x)+f(-x+2)=f(x)+f(2-x)

The value of that expression is 4 from the given.

In matrix form, the system is given by

[tex]\begin{bmatrix} -1 & 1 & -1 \\ 2 & -1 & 1 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -20 \\ 29 \\ 29 \end{bmatrix}[/tex]

I'll use G-J elimination. Consider the augmented matrix

[tex]\left[ \begin{array}{ccc|c} -1 & 1 & -1 & -20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right][/tex]

• Multiply through row 1 by -1.

[tex]\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right][/tex]

• Eliminate the entries in the first column of the second and third rows. Combine -2 (row 1) with row 2, and -3 (row 1) with row 3.

[tex]\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 5 & -2 & -31 \end{array} \right][/tex]

• Eliminate the entry in the second column of the third row. Combine -5 (row 2) with row 3.

[tex]\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 3 & 24 \end{array} \right][/tex]

• Multiply row 3 by 1/3.

[tex]\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right][/tex]

• Eliminate the entry in the third column of the second row. Combine row 2 with row 3.

[tex]\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right][/tex]

• Eliminate the entries in the second and third columns of the first row. Combine row 1 with row 2 and -1 (row 3).

[tex]\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 9 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right][/tex]

Then the solution to the system is

[tex]\boxed{x=9, y=-3, z=8}[/tex]

If you want to use G elimination and substitution, you'd stop at the step with the augmented matrix

[tex]\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right][/tex]

The third row tells us that [tex]z=8[/tex]. Then in the second row,

[tex]y-z = -11 \implies y=-11 + 8 = -3[/tex]

and in the first row,

[tex]x-y+z=20 \implies x=20 + (-3) - 8 = 9[/tex]

Answer:F to the power of 2,N to the power of 2,—25,f to the power of 2

Step-by-step explanation: i really hope under stand

The resulting value of the function when the values of t is 2 and 6 are 3.5 and 6.5 respectively

Functions are expressions in form of variables. Given the following quadratic equation expressed as:

f(t) = 0.25t^2 − 0.5t + 3.5

We can find the equivalent value of the function when the value of t a re 2 and 6.

If the value of t is 2. hence;

f(2) =  0.25(2)^2 − 0.5(2) + 3.5

f(2) = 1 - 1 + 3.5

f(2) = 3.5

If the value of t is 6, hence;

f(6) = 0.25(6)^2 − 0.5(6) + 3.5

f(6) = 9 - 3 + 3.5

f(6) = 6.5

Hence the resulting value of the function when the values of t is 2 and 6 are 3.5 and 6.5 respectively

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Answer:

[tex]x^2+y^2=64[/tex]

Step-by-step explanation:

So the first important thing in solving this problem, is identifying what the major and minor axis are. The major axis is the bigger one, and we want to find on which axis it is.

So by simply looking at the eclipse, you can see that it's larger on the horizontal axis, so the major axis is on the horizontal axis, and the minor axis is on the vertical axis.

This means the equation will be expressed as:

[tex]\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1[/tex]

In this equation the major axis, has a length of 2a, and the minor axis has a length of 2b. It's also important to note that (h, k) is the middle of the eclipse, but in the equation you provided, there is no subtraction, so it's just 0, meaning the center is at the origin (0, 0)

So now let's solve for a, and b. I'll look at each individual fraction separately.

[tex]\frac{x^2}{64}[/tex]

The denominator is equal to the value of a^2, so we simply take the square root of this to get the equation: a=8

[tex]\frac{y^2}{36}[/tex]

The denominator is equal to the value of b^2, so we simply take the square root of this to get the equation: b=6

So by just looking at the graph the larger circle appears to have a denominator that is equal to the length of the major axis. The major axis is equal to 2a, and since we know the value of a (8), we simply multiply this by 2 to get a length of 16. This was a bit redundant to do, since in the equation of a circle we need the radius, and the radius is just half the diameter, so now we divide this 16 by 2 to get a radius of 8.

The circle also appears to be centered at the origin so the equation will have (x-0)^2 and (y-0)^2 which is just x^2 and y^2

Plugging in all the values we get the equation:

[tex]x^2+y^2=64[/tex]

[tex] {\qquad\qquad\huge\underline{{\sf Answer}}} [/tex]

Here we go ~

The ellipse shown here is a horizontal ellipse that has major axis parallel to x - axis and minor axis parallel to y - axis. Length of major axis = 2a, and that of minor axis = 2b.

And it has it's centre on origin, it's equation can be written as :

[tex]\qquad \sf  \dashrightarrow \: \cfrac{ {x}^{2} }{ {a}^{2} } + \cfrac{ {y}^{2} }{ {b}^{2} } = 1[/tex]

so, let's equate given equation with the standard equation ~

[tex]\qquad \sf  \dashrightarrow \: \cfrac{ {x}^{2} }{64} + \cfrac{ {y}^{2} }{36} = 1[/tex]

so we get :

As we know, length of major axis is : 2a = 2 × 8 = 16 units

and, the larger circle has diameter = 2a = 16 units

so, it's radius = 8 units ~

Now, let's write the equation of circle with origin as centre and radius = 8 units

[tex]\qquad \sf  \dashrightarrow \: {(x - h)}^{2} + (y - k) {}^{2} = {r}^{2} [/tex]

[ h = 0, k = 0, since circle has centre at origin ]

[tex]\qquad \sf  \dashrightarrow \: {x}^{2} + {y}^{2} = 64[/tex]