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3 years ago

6

Simplify 16x^8y^5= 4x^5y

1 answer:

frosja888 [35]3 years ago

3 0

May you Break it down please

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E.) What is the surface area?and volume of sphere<br><br><br> √8 mm

CaHeK987 [17]

Answer:

v=4/3 pie r ³ you would put 8 in

Step-by-step explanation:

the formal

8 0

3 years ago

I don’t get how to find the base the height the side and the length

iren2701 [21]

Answer:

Well, the best way to think about it is base=bottom and then the height is the middle dotted line. The side lengths are the two sides of the triangle, not including the bottom. The two side lengths are 13 and the height is 12. I hope this helped a bit, if not then just let me know.

Step-by-step explanation:

7 0

3 years ago

Find the area of A parallelogram with a base of 12 inches and<br> height of 5 inches.

k0ka [10]

Step-by-step explanation:

Area=Base×Height

=12×5

=60

7 0

3 years ago

The original price of a computer was reduced by 225 dollars. The new, discounted price is 1274 dollars. Use p as your variable n

meriva

Answer:

1499 dollars

Step-by-step explanation:

Original price: p

p - 225 = 1274

add 225 to both sides

p = 1499

6 0

4 years ago

The sum of two numbers is 12, their product is 96. Compute these two numbers. Explain.

Anni [7]

Answer:

The numbers are

$6+2\sqrt{15}i$ and $6-2\sqrt{15}i$

Step-by-step explanation:

Let

x and y -----> the numbers

we know that

$x+y=12$ -----> $y=12-x$ ------> equation A

$xy=96$ ----> equation B

substitute equation A in equation B and solve for x

$x(12-x)=96\\12x-x^{2}=96\\x^{2} -12x+96=0$

Solve the quadratic equation

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^{2} -12x+96=0$

so

$a=1\\b=-12\\c=96$

substitute

$x=\frac{-(-12)(+/-)\sqrt{-12^{2}-4(1)(96)}} {2(1)}$

$x=\frac{12(+/-)\sqrt{-240}} {2}$

Remember that

$i^{2}=\sqrt{-1}$

$x=\frac{12(+/-)\sqrt{240}i} {2}$

$x=\frac{12(+/-)4\sqrt{15}i} {2}$

Simplify

$x=6(+/-)2\sqrt{15}i$

$x1=6+2\sqrt{15}i$

$x2=6-2\sqrt{15}i$

we have two solutions

<u><em>Find the value of y for the first solution</em></u>

For $x1=6+2\sqrt{15}i$

$y=12-x$

substitute

$y1=12-(6+2\sqrt{15}i)$

$y1=6-2\sqrt{15}i$

<u><em>Find the value of y for the second solution</em></u>

For $x2=6-2\sqrt{15}i$

$y2=12-x$

substitute

$y2=12-(6-2\sqrt{15}i)$

$y2=6+2\sqrt{15}i$

therefore

The numbers are

$6+2\sqrt{15}i$ and $6-2\sqrt{15}i$

8 0

4 years ago