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lys-0071 [83]
3 years ago
15

Ali drove 567 miles in 9 hours. At the same rate, how long would it take him to drive 441 miles?

Mathematics
1 answer:
adell [148]3 years ago
5 0

Answer:

7 hours

Step-by-step explanation:

567/9 = 63

441/63 = 7

It would take him 7 hours to drive 441 miles.

Hope this helps

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3 years ago
$32.00for a 14 2/9 km taxi ride. What is the cost per kilometer
KonstantinChe [14]

\begin{array}{ccll} \$&km\\ \cline{1-2} 32&14\frac{2}{9}\\[1em] x&1 \end{array}\implies \cfrac{32}{x}=\cfrac{14\frac{2}{9}}{1}\implies \cfrac{32}{x}=14\frac{2}{9}\implies \cfrac{32}{x}=\cfrac{14\cdot 9+2}{9} \\\\\\ \cfrac{32}{x}=\cfrac{128}{9}\implies 288=128x\implies \cfrac{288}{128}=x\implies \stackrel{~\hfill \textit{2 bucks and 25 cents}}{\cfrac{9}{4}=x\implies 2\frac{1}{4}}=x

4 0
2 years ago
I WILL GIVE YOU BRAINLIEST NEED HELP PLEASEE :DD
Shtirlitz [24]

The coordinates of the pre-image of point F' is (-2, 4)

<h3>How to determine the coordinates of the pre-image of point F'?</h3>

On the given graph, the location of point F' is given as:

F' = (4, -2)

The rule of reflection is given as

Reflection across line y = x

Mathematically, this is represented as

(x, y) = (y, x)

So, we have

F = (-2, 4)

Hence, the coordinates of the pre-image of point F' is (-2, 4)

Read more about transformation at:

brainly.com/question/4289712

#SPJ1

8 0
2 years ago
The circumference of a sphere was measured to be 76 cm with a possible error of 0.5 cm. (a) Use differentials to estimate the ma
andreev551 [17]

The maximum error in the calculated surface area is 24.19cm² and the relative error is 0.0132.

Given that the circumference of a sphere is 76cm and error is 0.5cm.

The formula of the surface area of a sphere is A=4πr².

Differentiate both sides with respect to r and get

dA÷dr=2×4πr

dA÷dr=8πr

dA=8πr×dr

The circumference of a sphere is C=2πr.

From above the find the value of r is

r=C÷(2π)

By using the error in circumference relation to error in radius by:

Differentiate both sides with respect to r as

dr÷dr=dC÷(2πdr)

1=dC÷(2πdr)

dr=dC÷(2π)

The maximum error in surface area is simplified as:

Substitute the value of dr in dA as

dA=8πr×(dC÷(2π))

Cancel π from both numerator and denominator and simplify it

dA=4rdC

Substitute the value of r=C÷(2π) in above and get

dA=4dC×(C÷2π)

dA=(2CdC)÷π

Here, C=76cm and dC=0.5cm.

Substitute this in above as

dA=(2×76×0.5)÷π

dA=76÷π

dA=24.19cm².

Find relative error as the relative error is between the value of the Area and the maximum error, therefore:

\begin{aligned}\frac{dA}{A}&=\frac{8\pi rdr}{4\pi r^2}\\ \frac{dA}{A}&=\frac{2dr}{r}\end

As above its found that r=C÷(2π) and r=dC÷(2π).

Substitute this in the above

\begin{aligned}\frac{dA}{A}&=\frac{\frac{2dC}{2\pi}}{\frac{C}{2\pi}}\\ &=\frac{2dC}{C}\\ &=\frac{2\times 0.5}{76}\\ &=0.0132\end

Hence, the maximum error in the calculated surface area with the circumference of a sphere was measured to be 76 cm with a possible error of 0.5 cm is 24.19cm² and the relative error is 0.0132.

Learn about relative error from here brainly.com/question/13106593

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3 0
2 years ago
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