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Kamila [148]
3 years ago
10

Rant Time

Mathematics
2 answers:
san4es73 [151]3 years ago
6 0

Answer: I would hit him over the head with a pan. <3

Step-by-step explanation: get a pan. hit him. hide the body. report the body if you can. Hope this helps <3

Lena [83]3 years ago
6 0

Answer:

Really meanly, break up.

lol.

Tell me if I gave you the right choice.

youBY THE WAY, I will NOT be responsabile for your depression,

and I will NOT be paying for your therapy fees.

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Miguel drove 737 miles in 11 hours.
Dmitriy789 [7]

7 hours , to find the amount of miles in one hour you do 737 divided by 11. Then you take that (67) and divide it by 469

4 0
3 years ago
What is the complete factorization of the polynomial below?<br> x3 + x2 + 9x+9
PIT_PIT [208]

Answer:

(x+1)(x-3i) (x+3i)

Step-by-step explanation:

x^3 + x^2 + 9x+9

We will use factor by grouping

Factor out x^2 from the first group and 9 from the second group

x^3 + x^2      + 9x+9

x^2( x+1)     + 9(x+1)

Factor out (x+1)

(x+1) ((x^2+9)

Now factor x^2+9

x^2 -(3i)^2 = (x-3i) (x+3i)

Replacing this

(x+1)(x-3i) (x+3i)

3 0
3 years ago
What is the equation of the following line written in general form? (The y-intercept is -1.)
Vedmedyk [2.9K]
~~~~~~~~~2x- y - 1 = 0 ~~~~~~~~~~
4 0
3 years ago
I need help on number 29
choli [55]

x = All real numbers or 0.

First you should distribute both sides of the equation. This gives you 42x+42=42x+42.

Then, subtract 42 from both sides. This gives you 42x=42x.

Now, divide 42 on both sides. This gives you x=0 or all real numbers.

7 0
3 years ago
Read 2 more answers
Help me solve this question
OLEGan [10]

Consider the first term on the left side.

\displaystyle \frac{2\sin{A}}{\cos{3A}}\\\\=\frac{2\sin{A}\cos{A}}{\cos{3A}\cos{A}}\qquad\text{multiply by cos(A)/cos(A)}\\\\=\frac{\sin{2A}}{\cos{3A}\cos{A}}\qquad\text{double angle formula for sin}\\\\=\frac{\sin{(2A-A)}}{\cos{3A}\cos{A}}=\frac{\sin{3A}\cos{A}-\cos{3A}\sin{A}}{\cos{3A}\cos{A}}\\\\=\tan{3A}-\tan{A}

Then the sum of the three terms on the left is

\left(\tan{3A}-\tan{A}\right)+\left(\tan{9A}-\tan{3A}\right)+\left(\tan{27A}-\tan{9A}\right)\\\\=\tan{27A}-\tan{A}\qquad\text{Q.E.D.}

4 0
4 years ago
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