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3 years ago

Given a polynomial f(x), if (x + 5) is a factor, what else must be true?

1 answer:

4 0

Rule: If x-k is a factor of p(x), then p(k) = 0. This is a special case of the remainder theorem.

Using that rule, we can see that x+5 is really x-(-5) so k = -5.

Therefore,

p(k) = 0
p(-5) = 0

so the answer is choice D.

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3 years ago

The surface area of a cylinder is increasing at a rate of 9 pi square meters per hour. The height of the cylinder is fixed at 3

Alekssandra [29.7K]

Answer:

$9\pi \text{ cubic meters per hour}$

Step-by-step explanation:

Since, the surface area of a cylinder,

$A= 2\pi r^2 + 2\pi rh$ ................(1)

Where,

r = radius,

h = height,

If $A= 36\pi\text{ square meters}, h = 3\text{ meters}$

$36\pi = 2\pi r^2 + 2\pi r(3)$

$18 = r^2 + 3r$

$\implies r^2 + 3r - 18=0$

$r^2 + 6r - 3r - 18 = 0$ ( by middle term splitting )

$r(r+6)-3(r+6)=0$

$(r-3)(r+6)=0$

By zero product property,

r = 3 or r = - 6 ( not possible )

Thus, radius, r = 3 meters,

Now, differentiating equation (1) with respect to t ( time ),

$\frac{dA}{dt}= 4\pi r\frac{dr}{dt} +2\pi(r\frac{dh}{dt} + h\frac{dr}{dt})$

∵ h = constant, ⇒ dh/dt = 0,

$\frac{dA}{dt} = 4\pi r \frac{dr}{dt} +2\pi h \frac{dr}{dt}$

We have, $\frac{dA}{dt}=9\pi\text{ square meters per hour}, r = h = 3\text{ meters}$

$9\pi = 4\pi (3) \frac{dr}{dt}+2\pi (3)\frac{dr}{dt}$

$9\pi = (12\pi + 6\pi )\frac{dr}{dt}$

$9\pi = 18\pi \frac{dr}{dt}$

$\implies \frac{dr}{dt} =\frac{1}{2}\text{ meter per hour}$

Now,

Volume of a cylinder,

$V=\pi r^2 h$

Differentiating w. r. t. t,

$\frac{dV}{dt}=\pi ( r^2 \frac{dh}{dt}+h(2r)\frac{dr}{dt})=\pi ((3)(6) (\frac{1}{2})) = 9\pi \text{ cubic meters per hour}$

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3 years ago

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3 years ago

A radioactive substance decays exponentially: The mass at time t is m(t) = m(0)ekt, where m(0) is the initial mass and k is a ne

velikii [3]

Answer:

M = 1/0.000121 = 8264.5 years

Step-by-step explanation:

M = − k ∫∞₀ teᵏᵗdt

To obtain this mean life, we'll use integration by parts to integrate the function ∫ teᵏᵗdt

∫udv = uv - ∫ vdu

u = t

du/dt = 1

du = dt

∫ dv = ∫ eᵏᵗdt

v = eᵏᵗ/k

∫udv = ∫ teᵏᵗdt

uv = teᵏᵗ/k

∫ vdu = eᵏᵗ/k

∫ teᵏᵗdt = (teᵏᵗ/k) - ∫eᵏᵗ/k

But, ∫eᵏᵗ/k = (1/k) ∫eᵏᵗ = (1/k²) eᵏᵗ = eᵏᵗ/k²

∫ teᵏᵗdt = (teᵏᵗ/k) - eᵏᵗ/k²

The rest of the calculation is done on paper in the image attached to this question

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