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2 years ago

6

olivia has a shoebox shaped like a rectangular prism. She decorates 4 faces of the box, leaving the top and bottom without decor

ations. The unshaded parts of the model and the diagram below show the parts she decorates. What is the total area olivia decorates

2 answers:

Art [367]2 years ago

8 0

Answer:

440 inches

Step-by-step explanation:

bruh thats too much work

marishachu [46]2 years ago

7 0

Answer:

440

Step-by-step explanation:

I already took the test g

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The 3rd degree Taylor polynomial for cos(x) centered at a = π 2 is given by, cos(x) = − (x − π/2) + 1/6 (x − π/2)3 + R3(x). Usin

Otrada [13]

Answer:

The cosine of 86º is approximately 0.06976.

Step-by-step explanation:

The third degree Taylor polynomial for the cosine function centered at $a = \frac{\pi}{2}$ is:

$\cos x \approx -\left(x-\frac{\pi}{2} \right)+\frac{1}{6}\cdot \left(x-\frac{\pi}{2} \right)^{3}$

The value of 86º in radians is:

$86^{\circ} = \frac{86^{\circ}}{180^{\circ}}\times \pi$

$86^{\circ} = \frac{43}{90}\pi\,rad$

Then, the cosine of 86º is:

$\cos 86^{\circ} \approx -\left(\frac{43}{90}\pi-\frac{\pi}{2}\right)+\frac{1}{6}\cdot \left(\frac{43}{90}\pi-\frac{\pi}{2}\right)^{3}$

$\cos 86^{\circ} \approx 0.06976$

The cosine of 86º is approximately 0.06976.

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2 years ago

Match each pair of points to the equation of the line that is parallel to the line passing through the points.

Paraphin [41]

we know that

If two lines are parallel, then, their slopes are equal.

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we will proceed to calculate the slope in each case, to determine the solution of the problem

<u>Case A)</u> Point $B(5,2)\ C(7,-5)$

Find the slope BC

Substitute the values in the formula

$m=\frac{-5-2}{7-5}$

$m=\frac{-7}{2}$

$m=-3.5$

so

The equation $y=-3.5x-15$ is parallel to the line passing through the points $B(5,2)\ C(7,-5)$

therefore

<u>the answer Part A) is</u>

$B(5,2)\ C(7,-5)$ ------> $y=-3.5x-15$

<u>Case B)</u> Point $D(11,6)\ E(5,9)$

Find the slope DE

Substitute the values in the formula

$m=\frac{9-6}{5-11}$

$m=\frac{3}{-6}$

$m=-0.5$

so

The equation $y=-0.5x-3$ is parallel to the line passing through the points $D(11,6)\ E(5,9)$

therefore

<u>the answer Part B) is</u>

$D(11,6)\ E(5,9)$ ------> $y=-0.5x-3$

<u>Case C)</u> Point $F(-7,12)\ G(3,-8)$

Find the slope FG

Substitute the values in the formula

$m=\frac{-8-12}{3+7}$

$m=\frac{-20}{10}$

$m=-2$

so

Any linear equation with slope $m=-2$ will be parallel to the line passing through the points $F(-7,12)\ G(3,-8)$

<u>Case D)</u> Point $H(4,4)\ I(8,9)$

Find the slope HI

Substitute the values in the formula

$m=\frac{9-4}{8-4}$

$m=\frac{5}{4}$

$m=1.25$

so

The equation $y=1.25x+4$ is parallel to the line passing through the points $H(4,4)\ I(8,9)$

therefore

<u>the answer Part D) is</u>

$H(4,4)\ I(8,9)$ ------> $y=1.25x+4$

<u>Case E)</u> Point $J(7,2)\ K(-9,8)$

Find the slope JK

Substitute the values in the formula

$m=\frac{8-2}{-9-7}$

$m=\frac{6}{-16}$

$m=-0.375$

so

Any linear equation with slope $m=-0.375$ will be parallel to the line passing through the points $J(7,2)\ K(-9,8)$

<u>Case F)</u> Point $L(5,-7)\ M(4,-12)$

Find the slope LM

Substitute the values in the formula

$m=\frac{-12+7}{4-5}$

$m=\frac{-5}{-1}$

$m=5$

so

The equation $y=5x+19$ is parallel to the line passing through the points $L(5,-7)\ M(4,-12)$

therefore

<u>the answer Part F) is</u>

$L(5,-7)\ M(4,-12)$ ------> $y=5x+19$

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