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DENIUS [597]
2 years ago
12

Which expression is equivalent to 15+3x A) 3(5+x) B) 5(3+x) C) 3(5+3x) D) 5(3+3x)

Mathematics
2 answers:
4vir4ik [10]2 years ago
7 0

Answer:

3(5+x)

Step-by-step explanation:

15+3x

5*3 + 3*x

Factor out 3

3(5+x)

Elodia [21]2 years ago
6 0
15x+3=3 (5x+1) I have taken this question before
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The path of a projectile launched from a 26-ft-tall tower is modeled by the equation y = −16t2 + 64t + 26. What is the maximum h
BabaBlast [244]

Answer:

The maximum height of the projectile is 90 ft

Step-by-step explanation:

Here, we want to get the maximum height reached by the projectile

The answer here will be the y-coordinate value of the vertex form of the given equation

so firstly, we have to write the equation in the vertex form

We have this as;

y = -16t^2 + 64t + 26

That will be;

y = a(x-h)^2 + k

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where the vertex of the equation is;

(-h,k)

K

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8 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

where:

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

r=3 sec(y)

a=0

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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mixer [17]

Answer:      

x2-6=5x

Step-by-step explanation:

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2 years ago
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