Nothing that 5 miles is equal to 8 kilometers. Convert 20 miles to kilometers.

F(x)= (x-2)(x-3)(x-5) simplify the right side. What is the equation?

Serga [27]

Answer:

x³ - 10x² + 31x - 30

Step-by-step explanation:

Given

(x - 2)(x - 3)(x - 5) ← expand the second pair of factors using FOIL

= (x - 2)(x² - 8x + 15) ← distribute

= x³ - 8x² + 15x - 2x² + 16x - 30 ← collect like terms

= x³ - 10x² + 31x - 30

5 0

3 years ago

A food pantry distributes 10-ounce bags of flour. A supermarket donates nine 5-pound bags to the pantry. How many 10-ounce bags

DanielleElmas [232]

Given:
nine 5 pound bags
distributed into 10-ounce bags

1 pound = 16 ounces

9 x 5 pounds = 45 pounds

45 pounds * 16 ounces/lbs = 45 * 16 oz = 720 ounces

720 ounces / 10 ounce bags = 720/10 = 72 bags.

The workers at the food pantry will be able to make 72 bags.

3 0

3 years ago

I can't find the answer

Rudik [331]

Answer: what grade is this

Step-by-step explanation:

3 0

3 years ago

What is the absolute value of -250?

Andrei [34K]

Answer:

250

Step-by-step explanation:

Pretty much every time it asks for the abasloute value it just asks how much to get the zero on the number line. It will ALWAYS be positive. For example -5 will be 5 to get to zero on the numberline. Same thing with a positive 5.

6 0

3 years ago

NO LINKS OR FILES!

Archy [21]

(a) If the particle's position (measured with some unit) at time t is given by s(t), where

$s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}$

then the velocity at time t, v(t), is given by the derivative of s(t),

$v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}$

(b) The velocity after 3 seconds is

$v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}$

(c) The particle is at rest when its velocity is zero:

$\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}$

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

$\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|$

In interval notation, this happens for t in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt$

By definition of absolute value, we have

$|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)$

In part (d), we've shown that v(t) > 0 when -√11 < t < √11, so we split up the integral at t = √11 as

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt$

and by the fundamental theorem of calculus, since we know v(t) is the derivative of s(t), this reduces to

$s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}$

7 0

2 years ago