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levacccp [35]
3 years ago
11

Which of the binomials below is a factor of this trinomial 3x^2+18x+24

Mathematics
2 answers:
finlep [7]3 years ago
7 0
I got D ! Keep practicing
Bingel [31]3 years ago
6 0
3(x^2+6x+8)
3(x+4)(x+2)
The answer is D
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Enter the missing numbers separated by a comma and no spaces!
spayn [35]
Y = 1x + 2 is the answer therefore the missing number is 1,2
7 0
3 years ago
What is the value of the expression below when n =<br> 2.4?<br> An X 2 + 13
aniked [119]

Answer:

34 because if u add the to the y

Step-by-step explanation: it would give u 34

8 0
3 years ago
How do you do this?
zaharov [31]
4sin x=2sin x + √3
4sin x-2sinx=√3
2sin x=√3
sinx=√3/2

x=arcsin √3/2=π/3  + 2Kπ U  2π/3+2Kπ

Sol: π/3  + 2Kπ   U    2π/3+2Kπ ;       K∈Z

π/3+2Kπ=60º+360ºk    
2π/3+2Kπ=120º+360ºK


4 0
3 years ago
A square piece of paper is folded in half vertically. If the resulting figure has a perimeter of 12 cm what was the area of the
Slav-nsk [51]

Answer:

24cm

Step-by-step explanation:

6+6+6+6=24

12 is half so you should add 12 or multiply by 2

8 0
3 years ago
Lim x-0 (sin2xcsc3xsec2x)/x²cot²4x
sergij07 [2.7K]

By the definitions of cosecant, secant, and cotangent, we have

\dfrac{\sin2x\csc3x\sec2x}{x^2\cot^24x}=\dfrac{\sin2x\sin^24x}{x^2\sin3x\cos2x\cos^24x}

Then we rewrite the fraction as

\dfrac{\sin2x}{2x}\left(\dfrac{\sin4x}{4x}\right)^2\dfrac{3x}{\sin3x}\dfrac{32}{3\cos2x\cos^24x}

The reason for this is that we want to apply the well-known limit,

\displaystyle\lim_{x\to0}\frac{\sin ax}{ax}=\lim_{x\to0}\frac{ax}{\sin ax}=1

for a\neq0. So when we take the limit, we have

\displaystyle\lim_{x\to0}\cdots=\lim_{x\to0}\frac{\sin2x}{2x}\left(\lim_{x\to0}\frac{\sin4x}{4x}\right)^2\lim_{x\to0}\frac{3x}{\sin3x}\lim_{x\to0}\frac{32}3\cos2x\cos^24x}

=1\cdot1^2\cdot1\cdot\dfrac{32}3=\dfrac{32}3

8 0
3 years ago
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