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3 years ago

Could anyone solve this?? Make m the subject of the formula 6m²=k Anyone??

1 answer:

3 0

Simplifying 6m2 = k Solving 6m2 = k Solving for variable 'm'. Move all terms containing m to the left, all other terms to the right. Divide each side by '6'. m2 = 0.1666666667k Simplifying m2 = 0.1666666667k Reorder the terms: -0.1666666667k + m2 = 0.1666666667k + -0.1666666667k Combine like terms: 0.1666666667k + -0.1666666667k = 0.0000000000 -0.1666666667k + m2 = 0.0000000000 The solution to this equation can't be determined for me

sorry :)

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3 years ago

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2 years ago

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Answer:

A) Each pack of folders has a volume of 60 cubic inches.

B) The box has a volume of about 720 cubic inches

D) If the box help 20 packs of folders, it would have a volume of about 1,200 cubic inches.

Step-by-step explanation:

Verify each statement

A) Each pack of folders has a volume of 60 cubic inches.

The statement is True

Because

The volume of each pack of folders is equal to

$V=(5)(12)(1)=60\ in^{3}$

B) The box has a volume of about 720 cubic inches

The statement is True

Because

The volume of the box is equal to the volume of one pack of folders multiplied by 12

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C) If the box held 15 packs of folders, it would have a volume of about 1,200 cubic inches

The statement is False

Because

Applying proportion

$\frac{12}{720}\frac{packs}{in^{3}}=\frac{15}{x}\frac{packs}{in^{3}}\\ \\x=720*15/12\\ \\x=900\ in^{3}$

$900\ in^{3}\neq 1,200\ in^{3}$

D) If the box help 20 packs of folders, it would have a volume of about 1,200 cubic inches.

The statement is True

Because

Applying proportion

$\frac{12}{720}\frac{packs}{in^{3}}=\frac{20}{x}\frac{packs}{in^{3}}\\ \\x=720*20/12\\ \\x=1,200\ in^{3}$

$1,200\ in^{3}= 1,200\ in^{3}$

E) Each pack of folders has a volume of 24 cubic inches.

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Because

The volume of each pack of folders is equal to

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3 years ago

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3 years ago

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Find the midpoint of the line segment with endpoints at the given coordinates (-6,6) and (-3,-9)

JulsSmile [24]

The midpoint of the line segment with endpoints at the given coordinates (-6,6) and (-3,-9) is $\left(\frac{-9}{2}, \frac{-3}{2}\right)$

Solution:

Given, two points are (-6, 6) and (-3, -9)

We have to find the midpoint of the segment formed by the given points.

The midpoint of a segment formed by $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { and }\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$ is given by:

$\text { Mid point } \mathrm{m}=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

$\text { Here in our problem, } x_{1}=-6, y_{1}=6, x_{2}=-3 \text { and } y_{2}=-9$

Plugging in the values in formula, we get,

$\begin{array}{l}{m=\left(\frac{-6+(-3)}{2}, \frac{6+(-9)}{2}\right)=\left(\frac{-6-3}{2}, \frac{6-9}{2}\right)} \\\\ {=\left(\frac{-9}{2}, \frac{-3}{2}\right)}\end{array}$

Hence, the midpoint of the segment is $\left(\frac{-9}{2}, \frac{-3}{2}\right)$

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3 years ago