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3 years ago

I need help and have to show my work

Mathematics

1 answer:

sweet [91]3 years ago

6 0

M+4=17
m+4-4=17
m=17-4
m=14
you have to get m alone so you subtract the 4 on both sides then when m is alone you subtract 17-4 and you get your answer

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The mass of the cylinder is 50000 g find the density of the cylinder to the nearest tenth

vichka [17]

Please consider the diagram of cylinder.

We have been given that the mass of the cylinder is 50000 g. We are asked to find the density of the cylinder.

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$

Let us find the volume of our given cylinder.

$V=\pi r^2 h$ , where

r = Radius,

h = Height.

We know that radius is half the diameter, so radius of given cylinder would be $\frac{28}{2}=14$ mm.

$V=\pi\cdot(14 \text{ mm})^2\cdot 30\text{ mm}$

$V=\pi\cdot 196\text{ mm}^2\cdot 30\text{ mm}$

$V=\pi\cdot 5880\text{ mm}^3$

$V=18472.5648031\text{ mm}^3$

$\text{Density}=\frac{50000\text{ g}}{18472.5648031\text{ mm}^3}$

$\text{Density}=2.7067167192510906\cdot \frac{\text{g}}{\text{ mm}^3}$

$\text{Density}\approx2.7\frac{\text{g}}{\text{ mm}^3}$

Therefore, the density of the cylinder is approximately 2.7 gram per cubic mm.

6 0

3 years ago

Ehich is bigger 3kl or 3000l

Yuliya22 [10]

1000L = 1 kL
3(1000L) = 3000 L
They are equal.

3 0

4 years ago

Find the taylor series for f(x centered at the given value of

Lady_Fox [76]

To find the Taylor series for f(x) = ln(x) centering at 5, we need to observe the pattern for the first four derivatives of f(x). From there, we can create a general equation for f(n). Starting with f(x), we have

$f(x) = ln(x) \\ f^{1}(x) = \frac{1}{x} \\ f^{2}(x) = -\frac{1}{x^{2}} \\ f^{3}(x) = \frac{2}{x^{3}} \\ f^{4}(x) = \frac{-6}{x^{4}}$
.
.
.
Since we need to have it centered at 5, we must take the value of f(5), and so on.

$f(5) = ln(5) \\ f^{1}(5) = \frac{1}{5} \\ f^{2}(5) = \frac{-1}{5^{2}} \\ f^{3}(5) = \frac{1(2)}{5^{3}} \\ f^{4}(5) = \frac{-1(2)(3)}{5^{4}}$
.
.
.
Following the pattern, we can see that for $f^{n}(x)$ ,

$f^{n}(x) = (-1)^{n-1} \frac{1(2)(3)...(n-1)}{5^{n}} \\ f^{n}(x) = (-1)^{n-1} \frac{(n-1)!}{5^{n}}$

This applies for $n\geq 1$ . Expressing f(x) in summation, we have

$\sum_{n=0}^{\infty} \frac{f^{n}(5)}{n!} (x-5)^{n}$

Combining ln2 with the rest of series, we have

$f(x) = ln2 + \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(n-1)!}{(n!)(5^{n})} (x-5)^{n}$
<span>
Answer: </span> $f(x) = ln2 + \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(n-1)!}{(n!)(5^{n})} (x-5)^{n}$

3 0

3 years ago

Greg is saving to buy a tablet that costs $235. He already has $25 saved up. Each week, Greg earns $15 babysitting and $25 as a

Nikitich [7]

Answer:

235=30x-25

Step-by-step explanation:

235+25=30x-25+25

260=30x

260/30=30x/30

x=8.66 or 9 weeks

8 0

4 years ago

Find the value. Show step by step. 0.5 x 0.5 x 0.5=

Ivan

Answer:

0.125

Explanation:

0.5 x 0.5 x 0.5 = 0.125

3 0

3 years ago