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BlackZzzverrR [31]
3 years ago
12

Which expressions are equivalent to equivalent to (V84 164 ) ? Select all that apply.​

Mathematics
1 answer:
Masja [62]3 years ago
3 0

Answer:

simplifying this expression (√8^4 × √6^4)^3/2 = 110592

so 2^12 ×3^3= 110592

similar 2^9 ×6^3 =110592

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Simplify the expression x(2x + 3) + (x - 3)(x - 4)
Art [367]
<span>x(2x + 3) + (x - 3)(x - 4)
=2x^2 + 3x +x^2 -3x -4x + 12
=3x^2 - 4x  + 12</span>
6 0
3 years ago
Multiply the binomial and the trinomial. <br> (x-2)(x^2+2x+4)
malfutka [58]

Answer:

Step-by-step explanation:

To solve this problem, we need to multiple x and -2 by x^{2} + 2x + 4 and add the results together:

x(x^{2} + 2x + 4)

x^{3} + 2x^{2} + 4x

-2(x^{2} + 2x + 4)

-2x^{2} - 4x - 8

Adding the two together give the following:

(x^{3} + 2x^{2} + 4x) + (-2x^{2} - 4x - 8)

x^{3} - 8

7 0
4 years ago
The function h(t) = −16t2 + 18t models the height, in feet, reached by a leopard t seconds after it jumps. What is the approxima
S_A_V [24]

9514 1404 393

Answer:

  5 1/16 ft

Step-by-step explanation:

  h(t) = -16t(t -18/16) . . . . put in intercept form

The function describes a parabola that opens downward. It has zeros at t=0 and t=9/8. The maximum height will be found at the vertex of the parabola, halfway between these zeros.

  f(9/16) = (-16)(9/16)² +18(9/16) = 81/16 = 5 1/16 . . . . feet

The approximate maximum height of the leopard is 5 1/16 feet.

5 0
3 years ago
Use the x-intercept method to find all real solutions of the equation. -9x^3-72x^2+36=3x^3+x^2-3x+8
larisa86 [58]

-9x^3-72x^2+36=3x^3+x^2-3x+8                     Add 9x^3 to both sides.

-72x^2 + 36 = 3x^3 + 9x^3 + x^2 - 3x + 8       Add 72x^2 to both sides

36 = 12x^3 +   73x^2 - 3x + 8                           Subtract 36 from both sides.

0 = 12x^3 + 73x^2 - 3x - 28      

It does factor, but it is not very nice.

(x + 6.06)(x - 6.09)(x + 0.632)

If there is any kind of error please report it in a note below.

6 0
3 years ago
Historically, the average time to service a customer complaint has been 3 days and the standard deviation has been 0.50 day. Man
kicyunya [14]
Oof I didn’t learn about this you can ask a class and go over it
6 0
3 years ago
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