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lina2011 [118]
3 years ago
12

Which are the roots of the quadratic function f(q) = q2 – 125? Check all that apply. q = 5 q = –5 q = 3 q = –3 q = 25 q = –25

Mathematics
2 answers:
AlladinOne [14]3 years ago
8 0

Answer:

q=5 sqrt 5 and q=-5 sqrt 5

Step-by-step explanation:

got it right on edge :)

slamgirl [31]3 years ago
7 0

Answer:

The given options are not roots of q(x)

Step-by-step explanation:

f(q)= q^2-125

To find the roots of the given quadratic equation

we replace f(q) with 0  and solve for q

0= q^2-125

Add 125 on both sides

125= q^2

take square root on both sides

q=+-5\sqrt{5}

q=+5\sqrt{5}  and  q=-5\sqrt{5}

The given options are not roots of q(x)

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

From the given information:

The domain D of integration in polar coordinates can be represented by:

D = {(r,θ)| 0 ≤ r ≤ 6, 0 ≤ θ ≤ 2π) &;

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y =\dfrac{\partial z}{\partial x} , x = \dfrac{\partial z}{\partial y}

Thus, the area of the surface is as follows:

\iint_D \sqrt{(\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2 +1 }\ dA = \iint_D \sqrt{(y)^2+(x)^2+1 } \ dA

= \iint_D \sqrt{x^2 +y^2 +1 } \ dA

= \int^{2 \pi}_{0} \int^{6}_{0} \ r  \sqrt{r^2 +1 } \ dr \ d \theta

=2 \pi \int^{6}_{0} \ r  \sqrt{r^2 +1 } \ dr

= 2 \pi \begin {bmatrix} \dfrac{1}{3}(r^2 +1) ^{^\dfrac{3}{2}} \end {bmatrix}^6_0

= 2 \pi \times \dfrac{1}{3}  \Bigg [ (37)^{3/2} - 1 \Bigg]

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3 years ago
Find the 7th term of the sequence
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Answer:

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4² = 16, 3rd term

.

.

.

4⁶ = 4096, 7th term

Answered by GAUTHMATH

8 0
3 years ago
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