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cluponka [151]
3 years ago
9

18x-16=-12x-4 what is the value of x?need help asap

Mathematics
2 answers:
Debora [2.8K]3 years ago
4 0
+12x +16
30x=12
12:30
x = 0,4
amm18123 years ago
3 0
18x - 16 = -12x - 4 ---> Add 12x to both sides
30x - 16 = -4 ---> Add 16 to both sides
30x = 12 ---> Divide both sides by 30

x = 0.4
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What is 24/30 written as a decimal
liubo4ka [24]

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Your brother is 1/2 your age. Your sister is 5 years older than your brother. Your sister is 15 years old. Write and solve an eq
nydimaria [60]

Answer:

a=2(s-5)

a(your age): 20

Step-by-step explanation:

a stands for your age

s stands for sister

a=2(s-5)

1-substitute 15 for s a=2(15-5)

2- a=2(10)

the 15-5 stands for sisters age minus five which gives you brothers age, 10. You multiply it by two because your brother is half your age.

3-a=20

check your work:

because your brother is 10 and you are 20, he is half your age (10/20 can be simplified to 1/2)

Hope this helped, if you have any other questions please lmk!

3 0
4 years ago
what are the vertex and x -intercepts of the graph of y=(x-4)(x+2)? Select one answer for the vertex and one for the x-intercept
kkurt [141]

Answer:

B&E

Step-by-step explanation:

Realized that I didn't actually "answer it" hahaha! Using a online calculator is always helpful in finding answers like these.

5 0
3 years ago
Find the volume of the solid generated when R​ (shaded region) is revolved about the given line. x=6−3sec y​, x=6​, y= π 3​, and
Dmitrij [34]

Answer:

V=9\pi\sqrt{3}

Step-by-step explanation:

In order to solve this problem we must start by graphing the given function and finding the differential area we will use to set our integral up. (See attached picture).

The formula we will use for this problem is the following:

V=\int\limits^b_a {\pi r^{2}} \, dy

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r=6-(6-3 sec(y))

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b=\frac{\pi}{3}

so the volume becomes:

V=\int\limits^\frac{\pi}{3}_0 {\pi (3 sec(y))^{2}} \, dy

This can be simplified to:

V=\int\limits^\frac{\pi}{3}_0 {9\pi sec^{2}(y)} \, dy

and the integral can be rewritten like this:

V=9\pi\int\limits^\frac{\pi}{3}_0 {sec^{2}(y)} \, dy

which is a standard integral so we solve it to:

V=9\pi[tan y]\limits^\frac{\pi}{3}_0

so we get:

V=9\pi[tan \frac{\pi}{3} - tan 0]

which yields:

V=9\pi\sqrt{3}]

6 0
3 years ago
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