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Setler79 [48]
3 years ago
6

1 point

Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
5 0

Answer:

(3⁰)×(4‐²)×(2³)

1×1/16×8

1/2

so th3 answer is option c.

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5.1x=-15.3 solve for x
LekaFEV [45]

Answer


x=-3


Hope it helps!!

3 0
3 years ago
Read 2 more answers
Is it possible to factor x cubed?
liberstina [14]
I dont think so.
factoring means make simpler
x cubed is already is simple so i dont think you cant. correct me if i am wrong
7 0
3 years ago
who does the number, of zeros in the product of 8 and 5000 compara to the number of zeros in the factors ? explain.
mezya [45]
If one of the numbers we multiply (factors) has zeros at the end, and the other isn't a fraction: all those zeros will stay in the product.

But there might be additional zeros if the other numbers  in the factors (the numbers which aren't 0) mupliply  to  "end" in zero and this is the case here:

8*5=40.

so the product will be 40 and the zeros of the 5000:

40 000

the number of zeros in the product will be bigger than the number of zeros in the factors if the non-zero parts of the fractions multiply to a number with 0 at the end.
4 0
3 years ago
Simplify cos^2 A/1+sin A
Stels [109]

Answer:

\displaystyle \frac{\cos^2(A)}{1+\sin(A)}=1-\sin(A)

Step-by-step explanation:

We want to simplify:

\displaystyle \frac{\cos^2(A)}{1+\sin(A)}

Recall the Pythagorean Identity:

\sin^2(A)+\cos^2(A)=1

So:

\cos^2(A)=1-\sin^2(A)

Substitute:

\displaystyle =\frac{1-\sin^2(A)}{1+\sin(A)}

Factor. We can use the difference of two squares pattern:

\displaystyle =\frac{\left(1-\sin(A))(1+\sin(A))}{1+\sin(A)}

Cancel. Hence:

\displaystyle \frac{\cos^2(A)}{1+\sin(A)}=1-\sin(A)

7 0
2 years ago
A cubic function with roots at 2, 0 and 3 containing the point at (5,-6)
natulia [17]

Answer:

Cubic equations and the nature of their roots

Just as a quadratic equation may have two real roots, so a cubic equation has possibly three. But unlike a quadratic equation which may have no real solution, a cubic equation always has at least one real root.

5 0
3 years ago
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