A rectangular prism is 4 inches long, 6 inches wide, and has a height of 5 inches. What is its volume?

Mathematics

1 answer:

Vadim26 [7]3 years ago

4 0

Answer:

Step-by-step explanation:

The volume of a rect. prism is V = (length)(width)(height).

Here the volume is V = (4 in)(6 in)(5 in) = 120 in³ or 120 in^3

Note: Please use " ^ " to indicate exponentiation: 120 in^3

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Find the equation of the line that passes through (0, -3) and is parallel to

Tresset [83]

Hey there!

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Answer:

$\green{\boxed{\red{\bold{\sf{y = \dfrac{7}{6}x - 3}}}}}$

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Explanation:

To find the equation of a line, we first have to determine its slope knowing that parallel lines have the same slope.

Let the line that we are trying to determine its equation be $\: \sf{d_1} \:$ and the line that is parallel to $\: \sf{d_1} \:$ be $\: \sf{d_2} \:$ .

$\sf{d_2} \:$ passes through the points (9 , 2) and (3 , -5) which means that we can find its slope using the slope formula:

$\sf{m = \dfrac{\Delta y}{\Delta x} = \dfrac{\green{y_2} - \orange{y_1}}{\red{x_2} - \blue{x_1 }}}$

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⇒Subtitute the values :

$\sf{(\overbrace{\blue{9}}^{\blue{x_1}}\: , \: \overbrace{\orange{2}}^{\orange{y_1}}) \: \: and \: \: (\overbrace{\red{3}}^{\red{x_2}} \: , \: \overbrace{\green{-5}}^{\green{y_2}} )}$

$\implies \sf{m = \dfrac{\Delta y}{\Delta x} = \dfrac{\green{-5} - \orange{2}}{\red{ \: \: 3} - \blue{9 }} = \dfrac{ - 7}{ - 6} = \boxed{ \bold{\dfrac{7}{6} }}}$

$\sf{\bold{The \: slope \: of \: both \: lines \: is \: \dfrac{7}{6}}}$ .

Assuming that we want to get the equation in Slope-Intercept Form, let's substitute m = 7/6:

Slope-Intercept Form:

$\sf{y = mx + b} \\ \sf{Where \: m \: is \: the \: slope \: of \: the \: line \: and \: b \: is \: the \: y-intercept.}$

$\implies \sf{y = \bold{\dfrac{7}{6}}x + b}$ $\\$

We know that the coordinates of the point (0 , -3) verify the equation since it is on the line $\: \sf{d_1} \:$ . Now, replace y with -3 and x with 0:

$\implies \sf{\overbrace{-3}^{y} = \dfrac{7}{8} \times \overbrace{0}^{x} + b} \\ \\ \implies \sf{-3 = 0 + b} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \sf{\boxed{\bold{b = -3}} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$