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3 years ago

Explain how two amounts of change can be the same but the percentage of change can be different

Mathematics

1 answer:

lys-0071 [83]3 years ago

8 0

The change from 10 to 11 or from 100 to 101 is 1 in each case.

In the first case, the percentage of change is 1/10 = 10%.

In the second case, the percentage of change is 1/100 = 1%.

The percentages of change are different because the bases from which the change is measure are different.

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What is the perimeter of the figure? <br> 229 4/9 ft<br> 229 2/3 ft<br> 230 2/3 ft <br> 231 ft

mote1985 [20]

The answer would be 230 2/3, because to get the perimeter you have to add all the sides together, 60 5/6 + 59 1/3 + 56 1/6 + 54 1/3 = 230 2/3

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3 years ago

a rectangular prism has base edge lengths of 6" and 8". The base edge lengths of a right rectangular prism are 8" and 6". The ba

FromTheMoon [43]

Answer:

Step-by-step explanation:

The heights of neither the rectangular prism nor the triangular prism are given, so we don't know the volume of either.

If h is the height of the rectangular prism, then its volume is

6×8×h = 48

if the height of the triangular prism is h/2, then its volume is

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So we know the volumes are the same -- but we don't know what that volume is.

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3 years ago

An accounting firm purchased a new digital camera that cost $2,500. The life of the camera is estimated to be 5 years. The total

lozanna [386]

2500-700 = 1800
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The annual depreciation of the camera is $360.

5 0

3 years ago

Consider the system of equations.

anzhelika [568]

Answer: You can multiply the top equation by -1 to eliminate the x variable.

And the solution is (2,4/3) in case you need it.

Step-by-step explanation:

2x + 3y = 8

2x + 6y = 12

If you multiply the upper equation or down equation by one, you will be able to eliminate the x variable.

-1( 2x + 3y) = -1(8) New equation: -2x -3y = -8.

Add the new equation you got by multiplying the top equation by -1 to the bottom equation.

Add them: -2x -3y = -8

2x + 6y = 12

3y = 4

y = 4/3

You can now input the value for y into the one of the equations and solve for x.

-2x - 3(4/3) = -8

-2x -4 = -8

+4 +4

-2x = -4

x = 2

8 0

3 years ago

What is the derivative of x times squaareo rot of x+ 6?

Dafna1 [17]

Hey there, hope I can help!

$\mathrm{Apply\:the\:Product\:Rule}: \left(f\cdot g\right)^'=f^'\cdot g+f\cdot g^'$
$f=x,\:g=\sqrt{x+6} \ \textgreater \ \frac{d}{dx}\left(x\right)\sqrt{x+6}+\frac{d}{dx}\left(\sqrt{x+6}\right)x \ \textgreater \ \frac{d}{dx}\left(x\right) \ \textgreater \ 1$

$\frac{d}{dx}\left(\sqrt{x+6}\right) \ \textgreater \ \mathrm{Apply\:the\:chain\:rule}: \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx} \ \textgreater \ =\sqrt{u},\:\:u=x+6$
$\frac{d}{du}\left(\sqrt{u}\right)\frac{d}{dx}\left(x+6\right)$

$\frac{d}{du}\left(\sqrt{u}\right) \ \textgreater \ \mathrm{Apply\:radical\:rule}: \sqrt{a}=a^{\frac{1}{2}} \ \textgreater \ \frac{d}{du}\left(u^{\frac{1}{2}}\right)$
$\mathrm{Apply\:the\:Power\:Rule}: \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1} \ \textgreater \ \frac{1}{2}u^{\frac{1}{2}-1} \ \textgreater \ Simplify \ \textgreater \ \frac{1}{2\sqrt{u}}$

$\frac{d}{dx}\left(x+6\right) \ \textgreater \ \mathrm{Apply\:the\:Sum/Difference\:Rule}: \left(f\pm g\right)^'=f^'\pm g^'$
$\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(6\right)$

$\frac{d}{dx}\left(x\right) \ \textgreater \ 1$
$\frac{d}{dx}\left(6\right) \ \textgreater \ 0$

$\frac{1}{2\sqrt{u}}\cdot \:1 \ \textgreater \ \mathrm{Substitute\:back}\:u=x+6 \ \textgreater \ \frac{1}{2\sqrt{x+6}}\cdot \:1 \ \textgreater \ Simplify \ \textgreater \ \frac{1}{2\sqrt{x+6}}$

$1\cdot \sqrt{x+6}+\frac{1}{2\sqrt{x+6}}x \ \textgreater \ Simplify$

$1\cdot \sqrt{x+6} \ \textgreater \ \sqrt{x+6}$
$\frac{1}{2\sqrt{x+6}}x \ \textgreater \ \frac{x}{2\sqrt{x+6}}$
$\sqrt{x+6}+\frac{x}{2\sqrt{x+6}}$

$\mathrm{Convert\:element\:to\:fraction}: \sqrt{x+6}=\frac{\sqrt{x+6}}{1} \ \textgreater \ \frac{x}{2\sqrt{x+6}}+\frac{\sqrt{x+6}}{1}$

Find the LCD
$2\sqrt{x+6} \ \textgreater \ \mathrm{Adjust\:Fractions\:based\:on\:the\:LCD} \ \textgreater \ \frac{x}{2\sqrt{x+6}}+\frac{\sqrt{x+6}\cdot \:2\sqrt{x+6}}{2\sqrt{x+6}}$

$Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions$
$\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{x+2\sqrt{x+6}\sqrt{x+6}}{2\sqrt{x+6}}$

$x+2\sqrt{x+6}\sqrt{x+6} \ \textgreater \ \mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c}$
$\sqrt{x+6}\sqrt{x+6}=\:\left(x+6\right)^{\frac{1}{2}+\frac{1}{2}}=\:\left(x+6\right)^1=\:x+6 \ \textgreater \ x+2\left(x+6\right)$
$\frac{x+2\left(x+6\right)}{2\sqrt{x+6}}$

$x+2\left(x+6\right) \ \textgreater \ 2\left(x+6\right) \ \textgreater \ 2\cdot \:x+2\cdot \:6 \ \textgreater \ 2x+12 \ \textgreater \ x+2x+12$
$3x+12$

Therefore the derivative of the given equation is
$\frac{3x+12}{2\sqrt{x+6}}$

Hope this helps!

8 0

3 years ago