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2 years ago

I need help with this 8(y+4)=7y+38

1 answer:

3 0

Distribute.

8(y+4)

8y+32= 7y+38

Subtract 32.

8y= 7y+6

Subtract 7y from both sides.

1y= 6

Divide 1 into both sides.

y=6

I hope this helped!

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How munch dose cups go in pints

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Consider the function below. (If an answer does not exist, enter DNE.) g(x) = 190 + 8x3 + x4 (a) Find the interval of increase.

galben [10]

Answer:

The interval of increase of g(x) is $(-6,+\infty)$ .

Step-by-step explanation:

The interval of increase occurs when first derivative of given function brings positive values. Let be $g(x) = 190 + 8 \cdot x^{3} + x^{4}$ , the first derivative of the function is:

$g'(x) = 24 \cdot x ^{2} + 4\cdot x^{3}$

$g'(x) = 4 \cdot x^{2}\cdot (6+x)$

The following condition must be met to define the interval of increase:

$4\cdot x^{2} \cdot (x+6) > 0$

The first term is always position due to the quadratic form, the second one is a first order polynomial and it is known that positive value is a product of two positive or negative values. Then, the second form must satisfy this:

$x + 6 > 0$

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$x > - 6$

Now, the solution to this expression in interval notation is: $(-6,+\infty)$

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3 years ago

Use the given transformation x=4u, y=3v to evaluate the integral. ∬r4x2 da, where r is the region bounded by the ellipse x216 y2

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The Jacobian for this transformation is

$J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix}$

with determinant $|J| = 12$ , hence the area element becomes

$dA = dx\,dy = 12 \, du\,dv$

Then the integral becomes

$\displaystyle \iint_{R'} 4x^2 \, dA = 768 \iint_R u^2 \, du \, dv$

where $R'$ is the unit circle,

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so that

$\displaystyle 768 \iint_R u^2 \, du \, dv = 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2 \, du \, dv$

Now you could evaluate the integral as-is, but it's really much easier to do if we convert to polar coordinates.

$\begin{cases} u = r\cos(\theta) \\ v = r\sin(\theta) \\ u^2+v^2 = r^2\\ du\,dv = r\,dr\,d\theta\end{cases}$

Then

$\displaystyle 768 \int_{-1}^1 \int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} u^2\,du\,dv = 768 \int_0^{2\pi} \int_0^1 (r\cos(\theta))^2 r\,dr\,d\theta \\\\ ~~~~~~~~~~~~ = 768 \left(\int_0^{2\pi} \cos^2(\theta)\,d\theta\right) \left(\int_0^1 r^3\,dr\right) = \boxed{192\pi}$

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1 year ago

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Mitra used 156 white tiles of 2x2 cm and 76 black tiles of 2x3 cm to form a pattern with an area of 1092 cm²

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Let x represent the number of white tiles and y represent the number of black tiles.

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Area of black tiles = 2 cm * 3 cm = 6 cm²

She used twice as many white tiles as black tiles. Hence:

x = 2y (1)

Also, the finished pattern was 1092 square centimeters in area. Hence:

4x + 6y = 1092 (2)

From both equations:

x = 156, y = 78

Mitra used 156 white tiles of 2x2 cm and 76 black tiles of 2x3 cm to form a pattern with an area of 1092 cm²

Find out more on equation at: brainly.com/question/2972832

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Answer:

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Step-by-step explanation:

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