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2 years ago

Help me solve these please

1 answer:

4 0

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7y-1=-1+9(y-9)
Add 1 on both sides.
7y=9(y-9)
Distribute.
7y=9y-81
Subtract 7y and add 81 on both sides.
2y=81
y=40.5
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4c+3=-12+c-1
Subtract 3 on both sides.
4c=-12-1-3+c
Subtract c on both sides.
3c=-16
c=-16/3
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-(k+1)-10=-5k+3
Add 10 on both sides.
-(k+1)=-5k+13
Distribute.
-k-1=-5k+13
Add 1 and add 5k on both sides.
4k=14
k=4/14
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-w-10=-9-2(w-8)
Distribute.
-w-10=-9-2w+16
Simplify.
-w-10=-2w+7
Add 2w and 10 on both sides.
w=17
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-2(n-3)-5=-5n+1
Distribute.
-2n+6-5=-5n+1
Simplify.
-2n+1=-5n+1
Add 5n and subtract 1 on both sides.
3n=0
n=0
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To make sure you got the right answer, you can check your work by plugging the result back into the question and seeing if both sides become equal. If both sides are equal, then the result is correct.
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For what value of k will the relation y = kx² + 8x + 8 have only one zero? (3 marks)

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2x^2 + 8x + 8

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2 years ago

The circumference of a sphere was measured to be 80 cm with a possible error of 0.5 cm. A) Use differentials to estimate the max

siniylev [52]

Answer:

A) The maximum error in the calculated surface area: $25cm^2$

Relative error: $0.013$

B) The maximum error in the calculated volume: $162cm^2$

Relative error: $0.019$

Step-by-step explanation:

A) The formula for the surface area is:

$A=4\pi r^2$

The measured value is the circumference which is equal to:

$C=2\pi r$

then the radius is:

$r=\frac{C}{2\pi}$

Substituting in the formula of the surface:

$A=4\pi(\frac{C}{2\pi})^2\\A=4\pi(\frac{C^2}{4\pi^2})\\A=\frac{C^2}{\pi}$

Using the formula to calculate the error:

$dy=f'(x)dx$

Where $x$ is the variable measured and $y$ is a function of $x$ ( $y=f(x)$ ).

$dA=f'(C)dC\\dA=\frac{2C^{(2-1)}}{\pi}dC\\dA=\frac{2C}{\pi}dC$

We have C=80cm and dC=0.5cm

$dA=\frac{2C}{\pi}dC\\dA=\frac{2(80)}{\pi}(0.5)\\dA=\frac{160}{\pi}(0.5)\\dA=50.9296(0.5)\\dA=25.4648\approx25cm^2$

The relative error is the maximum error divide by the total area. The total area is: $A=\frac{C^2}{\pi}=\frac{(80)^2}{\pi}=\frac{6400}{\pi}=2037.1833cm^2$

$\frac{dA}{A}=\frac{25.4648}{2037.1833} =0.0125\approx0.013$

B) The formula for the volume is:

$V=\frac{4}{3} \pi r^3$

Using $r=\frac{C}{2\pi}$

$V=\frac{4}{3} \pi r^3\\V=\frac{4}{3} \pi (\frac{C}{2\pi})^3\\V=\frac{4}{3} \pi (\frac{C^3}{8\pi^3})\\V=\frac{1}{3}(\frac{C^3}{2\pi^2})\\V=\frac{C^3}{6\pi^2}$

The maximum error is:

$dV=\frac{3C^{3-1}}{6\pi^2}dC\\dV=\frac{C^{2}}{2\pi^2}dC\\dV=\frac{(80)^{2}}{2\pi^2}(0.5)\\dV=\frac{6400}{2\pi^2}(0.5)\\dV=\frac{6400}{2\pi^2}(0.5)\\dV=(324.2278)(0.5)\\dV=162.1139\approx162cm^2$