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3 years ago

A chocolate bar is packaged in a box on the shape of a triangular prism.

Mathematics

1 answer:

BartSMP [9]3 years ago

8 0

Answer:

20 cm

Step-by-step explanation:

A prism's volume is found using the formula A = B*h where B is the are of the base. Here the volume v = 250 and the B = 12.5. Substitute the values and solve for h.

250 = 12.5*h

250/12.5 = h

20 = h

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Half of j minus 5 is the sum of k and 13

barxatty [35]

.5j - 5 = k + 13. Is that what you want?

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3 years ago

Solve for x: 1 < X + 3 < 4

sveta [45]

Answer:

for soln 1: X < -2; for soln 2: X < 1

Step-by-step explanation:

This is two set problem involving inequality. Therefore, you have to solve for each side of the inequality to know the values of X for each sides.

1 < X + 3 < 4

Soln 1. solve for the first side of the inequality

1 < X + 3

1 - 3 < X

-2 < X

X > -2

soln 2: solve for the second side of the inequality

X +3 < 4

X < 4 - 3

X < 1

Therefore, X > -2; X < 1

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2 years ago

Which of the following numbers is not the difference between two of the others?

luda_lava [24]

<span>A.1 - the difference between 5 and 6
B.7
C.6 - the difference between 7 and 1
D.5 - the difference between 7 and 2
E.2 - the difference between 7 and 5

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4 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%28W%5E%7B8%7D%20%29%5E%7B8%7D" id="TexFormula1" title="(W^{8} )^{8}" alt="(W^{8} )^{8}" align

MaRussiya [10]

Answer:

Step-by-step explanation:

$\left(w^8\right)^8\\$

$\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc}$

$\left(w^8\right)^8=w^{8\times\:8}\\=w^6^4$

4 0

3 years ago

Find an exact value.

Westkost [7]

Answer:

$\displaystyle \cos\left(-\frac{7\,\pi}{12}\right) = \frac{\sqrt{2} - \sqrt{6}}{4}$ .

Step-by-step explanation:

Convert the angle $\displaystyle \left(-\frac{7\, \pi}{12}\right)$ to degrees:

$\displaystyle \left(-\frac{7\, \pi}{12}\right) = \left(-\frac{7\, \pi}{12}\right) \times \frac{180^\circ}{\pi} = -105^\circ$ .

Note, that $\left(-105^\circ\right)$ is the sum of two common angles: $\left(-45^\circ\right)$ and $\left(-60^\circ\right)$ .

$\displaystyle \cos\left(-45^\circ\right) = \cos\left(45^\circ\right) = \frac{\sqrt{2}}{2}$ .
$\displaystyle \cos\left(-60^\circ\right) = \cos\left(60^\circ\right) = \frac{1}{2}$ .

$\displaystyle \sin\left(-45^\circ\right) = -\sin\left(45^\circ\right) = -\frac{\sqrt{2}}{2}$ .
$\displaystyle \sin\left(-60^\circ\right) = -\sin\left(60^\circ\right) = -\frac{\sqrt{3}}{2}$ .

By the sum-angle identity of cosine:

$\cos(A + B) = \cos(A)\cdot \cos(B) - \sin(A) \cdot \sin(B)$ .

Apply the sum formula for cosine to find the exact value of $\cos\left(-105^\circ \right)$ .

$\begin{aligned}\cos\left(-105^\circ \right) &= \cos\left(\left(-45^\circ\right) + \left(-60^\circ\right)\right) \\ &= \cos\left(-45^\circ\right) \cdot \cos\left(-60^\circ\right)\right) - \sin\left(-45^\circ\right) \cdot \sin\left(-60^\circ\right)\right) \\ &= \frac{\sqrt{2}}{2} \times \frac{1}{2} - \left(-\frac{\sqrt{2}}{2}\right)\times \left(-\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{2} - \sqrt{6}}{4}\end{aligned}$ .

$\displaystyle \left(-\frac{7\, \pi}{12}\right) = \left(-\frac{7\, \pi}{12}\right) \times \frac{180^\circ}{\pi} = -105^\circ$ . In other words, $\displaystyle \left(-\frac{7\, \pi}{12}\right)$ and $\left(-105^\circ\right)$ correspond to the same angle. Therefore, the cosine of $\displaystyle \left(-\frac{7\, \pi}{12}\right)\!$ would be equal to the cosine of $\left(-105^\circ\right)\!$ .

$\displaystyle \cos\left(-\frac{7\,\pi}{12}\right) = \cos\left(-105^\circ\right) = \frac{\sqrt{2} - \sqrt{6}}{4}$ .

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3 years ago