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3 years ago

One nickel weighs 5 g. Find how many nickels are in 7.0 kg of nickels

Mathematics

1 answer:

malfutka [58]3 years ago

4 0

Answer: there are 200 nickels in 1 kilogram of nickels

Step-by-step explanation:

For this case we must take into account the following conversion:

1Kg = 1000 g

Therefore, we can make the following rule of three:

5 grams ---------> 1 nickel

1000 grams ----> x

Clearing x we have:

x = (1000/5) * (1)

x = 200 nickels

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Kawai requires a loan of $33,000 to buy a used sport utility vehicle. His banks offers financing at 6.75% compounded monthly, fo

UkoKoshka [18]

Answer:

$49,126.87

Step-by-step explanation:

Given data

Principal= $33,000

rate= 6.75%

Time= 6 years

First, convert R as a percent to r as a decimal

r = R/100

r = 6.65/100

r = 0.0665 rate per year,

Then solve the equation for A

A = P(1 + r/n)^nt

A = 33,000.00(1 + 0.0665/12)^(12)(6)

A = 33,000.00(1 + 0.005541667)^(72)

A = $49,126.87

Hence the final amount is $49,126.87

7 0

2 years ago

3 regions are defined in the figure find the volume generated by rotating the given region about the specific line

anastassius [24]

The volume generated by rotating the given region $R_{3}$ about OC is $\frac{4}{g} \pi$

<h3>Washer method</h3>

Because the given region ( $R_{3}$ ) has a look like a washer, we will apply the washer method to find the volume generated by rotating the given region about the specific line.

solution

We first find the value of x and y

$y=2(x)^{\frac{1}{4} }$

$x=(\frac{y}{2} )^{4}$

$y=2x$

$x=\frac{y}{2}$

$\int\limits^a_b {\pi } \, (R_{o^{2} } - R_{i^{2} } ) dy$

$R_{o} = x = \frac{y}{2}$

$R_{i} = x= (\frac{y}{2}) ^{4}$

$a=0, b=2$

$v= \int\limits^2_o {\pi } \, [(\frac{y}{2})^{2} - ((\frac{y}{2}) ^{4} )^{2} ) dy$

$v= \pi \int\limits^2_o= [\frac{y^{2} }{4} - \frac{y^{8} }{2^{8} }} ] dy$

$v= \pi [\int\limits^2_o {\frac{y^{2} }{4} } \, dy - \int\limits^2_o {\frac{y}{2^{8} } ^{8} } \, dy ]$

$v=\pi [\frac{1}{4} \frac{y^{3} }{3} \int\limits^2_0 - \frac{1}{2^{8} } \frac{y^{g} }{g} \int\limits^2_o\\v= \pi [\frac{1}{12} (2^{3} -0)-\frac{1}{2^{8}*9 } (2^{g} -0)]\\v= \pi [\frac{2}{3} -\frac{2}{g} ]\\v= \frac{4}{g} \pi$