All categories

1 year ago

This is easy for other students in my class just not me lol

Mathematics

1 answer:

zavuch27 [327]1 year ago

8 0

Answer:

Step-by-step explanation:

Angle FGH = x

Angle FGJ = 25

Angle JGH = 40

Angle FGH = Angle FGJ + Angle JGH

x = 25 + 40

x = 65

You might be interested in

Kyle and Linda are married with two children at home and a mortgage. Kyle’s net pay per year is $32,000 and Linda’s is $48,000.

natka813 [3]

As their financial advisor,I would have them set aside more money for their savings and their retirement fund.Also I would advise them to put out a life insurance policy on linda since she doesn't have one so if something every happenes then kyle would have money to raise the kids since Linda makes more money.I would also gave them make college funds for their children since college is so expensive that way they wouldn't have to worry about it in the future.

I hope this helped. :)

4 0

2 years ago

Read 2 more answers

WILL GIVE BRAILIEST IF YOU ANSWER!! The MTC Theater holds 250 people, and last night’s performance sold out! Students under the

Alexandra [31]

Answer:

Adult Tickets: 190

16 and less: 60

Step-by-step explanation:

190 x 4.00 + 60 x 2.5 = 910

8 0

2 years ago

What is the slope for 12 34

NemiM [27]

We can't solve it with that information

8 0

2 years ago

The area enclosed by the graphs of y = 1/x, y = 1, and x = 3 is rotated about the line y = -1. Find the volume and show steps.

Natalija [7]

Answer:

$A=\pi\displaystyle\biggr[\frac{16}{3}-2\ln(|3|)\biggr]\approx9.8524$

Step-by-step explanation:

Use the Washer Method $A=\pi\displaystyle \int\limits_{a}^{b}{\bigr[R(x)^2-r(x)^2\bigr] \, dx$ where $R(x)$ is the outer radius and $r(x)$ is the inner radius.

If we sketch out the graph, we see that $y=1$ intersects points $(1,1)$ and $(3,1)$ , which will be our bounds of integration.

Here, our outer radius will be $R(x)=(-1-1)=-2$ and our inner radius will be $r(x)=-1-\frac{1}{x}$ .

Thus, we can compute the integral and find the volume:

$A=\pi\displaystyle\int\limits^{3}_{1} {(-2)^2-\biggr(-1-\frac{1}{x}\biggr)^2 } \, dx\\ \\A=\pi\displaystyle\int\limits^{3}_{1} {4-\biggr(1+\frac{2}{x}+\frac{1}{x^2} \biggr) } \, dx\\\\A=\pi\displaystyle\int\limits^{3}_{1} {4-1-\frac{2}{x}-\frac{1}{x^2}} \, dx\\\\A=\pi\displaystyle\int\limits^{3}_{1} {3-\frac{2}{x}-\frac{1}{x^2}} \, dx\\\\A=\pi\displaystyle\biggr[3x-2\ln(|x|)+\frac{1}{x}\biggr]\Biggr|_{1}^{3}\\$

$A=\pi\displaystyle\biggr[\biggr(3(3)-2\ln(|3|)+\frac{1}{3}\biggr)-\biggr(3(1)-2\ln(|1|)+\frac{1}{1}\biggr)\biggr]\\\\A=\pi\displaystyle\biggr[\biggr(9-2\ln(|3|)+\frac{1}{3}\biggr)-\biggr(3+1\biggr)\biggr]\\\\A=\pi\displaystyle\biggr[\biggr(\frac{28}{3}-2\ln(|3|)\biggr)-\biggr(4\biggr)\biggr]\\A=\pi\displaystyle\biggr[\frac{16}{3}-2\ln(|3|)\biggr]\\A\approx9.8524$