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Sever21 [200]
3 years ago
11

8c^3-27d^3 factored

Mathematics
1 answer:
Andreas93 [3]3 years ago
5 0
Hi Cinna919

8c³ - 27d³
There is no way we can factorize this, we can't even simplify it. As a result, it is impossible for us to factorize this.

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5. A boy owns 2 pairs of pants, 8 shirts, 6 ties, and 7 jackets. How many different outfits can he wear to school if he must wea
sergeinik [125]

Answer:

672

Step-by-step explanation:

6 0
3 years ago
Solve<br>3 tan 8 &lt;-1 for 0 5 0521.​
BlackZzzverrR [31]

Answer:

Substitute the given value into the function and evaluate.

Substitute the given value into the function and evaluate.

3  tan  ( 8 )  <  −  1  =  0.4216225  <  −  1

0  =  0.4216225  <  −  1

5  =  0.4216225  <  −  1

521  =  0.4216225  <  −  1

Step-by-step explanation:

3 0
3 years ago
Image of math question below
kari74 [83]
The second one is the correct answer

4 0
3 years ago
Read 2 more answers
NEED HELP ASAP
algol [13]

Answer:

A) car one is linear because it decreases the same amount every year.  

    car two decreases by the same ratio so it's exponential.  

B) for car one y = -6,000x + 38,000 (y is the value after x years, -6,000 is what it changes by every year, and 38,000 is the amount of year 0)

   for car two y = (44,000) (17/20)x-1 (y is the value after x years, 44,000 is the amount on year 0, and 17/20 is the ratio)

C) after 5 years car one will be 8,000

    after 5 years car two will be about 16,860

7 0
3 years ago
Which of the following is not a solution to the system of linear equations below?
ra1l [238]

Answer:

<h2>B. (-15, 12)</h2>

Step-by-step explanation:

\left\{\begin{array}{ccc}5y=3x+15&\text{subtract 3x from both sides}\\6x=10y-30&\text{subtract 10y from both sides}\end{array}\right\\\left\{\begin{array}{ccc}-3x+5y=15\\6x-10y=-30&\text{divide both sides by 2}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}-3x+5y=15\\3x-5y=-15\end{array}\right}\\.\qquad0=0\qquad\bold{TRUE}\\\\\bold{Infinitely\ many\ solutions}\\\\5y=3x+15\qquad\text{divide both sides by 5}\\\\y=\dfrac{3}{5}x+3\qquad(*)\\\\\text{Put the coordinates of the points to the equation (*)}

A.\ (5,\ 6)\\\\6=\dfrac{3}{5}(5)+3\\\\6=3+3\\\\6=6\qquad\bold{CORRECT}\\\\B.\ (-15,\ 12)\\\\12=\dfrac{3}{5}(-15)+3\\\\12=(3)(-3)+3\\\\12=-9+3\\\\12=-6\qquad\bold{FALSE}\\\\C.\ (0,\ 3)\\\\3=\dfrac{3}{5}(0)+3\\\\3=0+3\\\\3=3\qquad\bold{CORRECT}\\\\D.\ (-10,\ -3)\\\\-3=\dfrac{3}{5}(-10)+3\\\\-3=(3)(-2)+3\\\\-3=-6+3\\\\-3=-3\qquad\bold{CORRECT}

5 0
4 years ago
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