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3 years ago

Both cases?

Mathematics

1 answer:

pishuonlain [190]3 years ago

8 0

Answer:16.7 percent that what I got

Step-by-step explanation:

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-5x -10 >30 help Hehehehehe.

Mnenie [13.5K]

Answer:

x<-8

Step-by-step explanation:

-5x-10>30

(add 10 to both sides)

-5x-10+10>30+10

-5x>40

$\frac{-5x}{-1}$ < $\frac{40}{-1}$

(dividing through by -1 changes the sign of the inequality from "<" to ">")

5x<-40

(dividing through by 5)

$\frac{5x}{5}$ < $\frac{-40}{5\\}$

x<-8

4 0

3 years ago

Let f(x) = 4x – 5 and g(x) = 3x + 7. find f(x) + g(x) and state its domain.

Tema [17]

We are given the functions:

f(x) = 4 x – 5 ---> 1

g(x) = 3 x + 7 ---> 2

To find for the value of f(x) + g(x), all we have to do is to add equations 1 and 2:

f(x) + g(x) = 4 x – 5 + 3 x + 7

f(x) + g(x) = 7 x + 2 = y

In this case, for any real number value assign to x, we get a real number value of y. This is because the function is linear.

Therefore the domain of the function is all real numbers.

5 0

3 years ago

Rolling an even number and the coin landing on heads

iren [92.7K]

Answer:

$\frac{1}{4}$

Step-by-step explanation:

Rolling a even: $\frac{3}{6} -> \frac{1}{2}$

Coin on heads: $\frac{1}{2}$

$\frac{1}{2}$ × $\frac{1}{2} =\frac{1}{4}$

5 0

3 years ago

Read 2 more answers

Use the formula v = u + at to find the velocity when the initial velocity is 3 m/s the acceleration is 1.5 m/s the time is 7 s

vichka [17]

$answer \\ 13.5 \: m/s \\ given \\ initial \: velocity(u) = \: 3 \: m/s \\ acceleration(a) = 1.5 \: m/s ^2\\ time(t) = 7 \: second \\ now \\ v = u + at \\ or \: v = 3 + 1.5 \times 7 \\ \: or \: v = 3 + 10.5 \\ v = 13.5 \: m/s \\ hope \: it \: helps$

4 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