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tensa zangetsu [6.8K]
3 years ago
10

0.1 is 10% of what number?

Mathematics
2 answers:
stiv31 [10]3 years ago
7 0
Let's assign a variable that is the number when multiplied by 10% it is equal to 0.1.

Well let that be x.

Now we want to convert the 10% into a decimal. To convert a percent into a decimal, simply move the % sign two places to the left and make it into a decimal point.

When we say "of" in mathematics what we really mean is to multiply.

0.10 * x = 0.1

That is the equation we need to find the value of x for.

Divide both sides by 0.10

0.10 * x = 0.1

The 0.10 on the left side cancels out.

x = 1

So, the answer to this question is if we multiply 10% by 1 we will get 0.1.
sesenic [268]3 years ago
7 0
The answer is 1. ^that is a correct explanation
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Evaluate 4x3 + 2 for x = -1.
LekaFEV [45]
4 * -1 * 3 + 2 = -10
Sorry if I didn't do this right!
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Evaluate the given expression.<br><br> C(10,4)
Alex_Xolod [135]

Answer:

c(10) = 4

Step-by-step explanation:

4 0
3 years ago
Find the values of m and b that make the following function differentiable.
KengaRu [80]

f(x)=\begin{cases}x^2&\text{for }x\le2\\mx+b&\text{for }x>2\end{cases}

In order to be differentiable everywhere, f must first be continuous everywhere, which means the limits from either side as x\to2 must be the same and equal to f(2). By definition, f(2)=2^2=4, and

\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2}x^2=4

\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2}(mx+b)=2m+b

so we need to have 4m+b=4.

For f to be differentiable at x=2, the derivative needs to be continuous at x=2, i.e.

\displaystyle\lim_{x\to2^-}f'(x)=\lim_{x\to2^+}f'(x)

We then need to have

\displaystyle\lim_{x\to2}2x=\lim_{x\to2}m\implies\boxed{m=4}

Then

2m+b=4\implies\boxed{b=-4}

7 0
3 years ago
First make a substitution and then use integration by parts to evaluate the integral. (Use C for the constant of integration.) x
e-lub [12.9K]

Answer:

(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C

Step-by-step explanation:

Ok, so we start by setting the integral up. The integral we need to solve is:

\int x ln(5+x)dx

so according to the instructions of the problem, we need to start by using some substitution. The substitution will be done as follows:

U=5+x

du=dx

x=U-5

so when substituting the integral will look like this:

\int (U-5) ln(U)dU

now we can go ahead and integrate by parts, remember the integration by parts formula looks like this:

\int (pq')=pq-\int qp'

so we must define p, q, p' and q':

p=ln U

p'=\frac{1}{U}dU

q=\frac{U^{2}}{2}-5U

q'=U-5

and now we plug these into the formula:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int \frac{\frac{U^{2}}{2}-5U}{U}dU

Which simplifies to:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int (\frac{U}{2}-5)dU

Which solves to:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\frac{U^{2}}{4}+5U+C

so we can substitute U back, so we get:

\int xln(x+5)dU=(\frac{(x+5)^{2}}{2}-5(x+5))ln(x+5)-\frac{(x+5)^{2}}{4}+5(x+5)+C

and now we can simplify:

\int xln(x+5)dU=(\frac{x^{2}}{2}+5x+\frac{25}{2}-25-5x)ln(5+x)-\frac{x^{2}+10x+25}{4}+25+5x+C

\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}-\frac{5x}{2}-\frac{25}{4}+25+5x+C

\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C

notice how all the constants were combined into one big constant C.

7 0
3 years ago
Hi its my birthday was i correct ? if not please help
s344n2d4d5 [400]
ANSWER: The answer is c) -5 and a very happy birthday to you
7 0
2 years ago
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