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3 years ago

How to tell the different properties in algebra

Mathematics

1 answer:

Korvikt [17]3 years ago

5 0

There are about 10 properties that you need to remember.

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What is the volume of a cube with a side length of 3t^5units?<br>Cube V = s^3

balandron [24]

Answer:

$V=27t^{15}\ units^3$

Step-by-step explanation:

we know that

The volume of a cube is equal to

$V=s^{3}$

where

s is the side length of the cube

we have

$s=3t^5\ units$

substitute in the formula

$V=(3t^5)^{3}$

$V=(3^3)(t^5)^{3}$

Remember the power rule

$(a^m)^n=a^{m*n}$ ----> multiply the exponents

$V=(3^3)(t^{5*3})=27t^{15}\ units^3$

6 0

2 years ago

Read 2 more answers

What is the value of x in the equation 4(2x + 10) = 0? (1 point)

miskamm [114]

Answer:

x = -5

Explanation:

4(2x + 10) = 0

[ Simplify both sides of the equation ]

4(2x + 10) = 0

(4)(2x) + (4)(10) = 0 [Distribute]

8x + 40 = 0

[ Subtract 40 from both sides ]

8x + 40 − 40 = 0 − 40

8x = −40

[ Divide both sides by 8 ]

8x / 8 = −40 / 8

x = -5

4 0

2 years ago

Identify the "inside function" u = f(x) and the "outside function" y = g(u). Then find dy/dx using the Chain Rule.

skad [1K]

DfLet $f(x)=\sec x$ and $g(x)=\sqrt x$ . Then

$y=\sec\sqrt x=\sec(g(x))=f(g(x))=f\circ g(x)$

By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm du}\dfrac{\mathrm du}{\mathrm dx}$

where $u=g(x)=\sqrt x$ , so that $y=f(g(x))=f(u)=\sec u$ . We have

$\dfrac{\mathrm du}{\mathrm dx}=\dfrac1{2\sqrt x}$
$\dfrac{\mathrm d\sec u}{\mathrm du}=\sec u\tan u$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\sec u\tan u\dfrac{\mathrm dy}{\mathrm du}=\dfrac{\sec\sqrt x\,\tan\sqrt x}{2\sqrt x}$