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3 years ago

Factor the triomial x2-8x+15

1 answer:

3 0

X²-8x+15
Use the p and q method.
p*q=15
p+q=-8
p=-5, q=-3
Write in your factored form.
(x-5)(x-3)
So, (x-5)(x-3) is your answer.

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Can someone please help me with this question

zlopas [31]

Answer:

15 kilometers

Step-by-step explanation:

So we know:

For 4 days she runs 1500 meters each day.

For 3 days she runs 3 kilometers each day.

First off, lets convert 1500 meters into kilometers.

There are 1000 meters in a kilometer, so pluggin in 1500 meters:

1500/1000 = 1.5

So for 4 days she runs 1.5 kilometers each day.

Now, to find the total kilometers, we must find multiply the 1.5 kilometers by the 4 days, the 3 kilometers by the 3 days, then add those two together.

So lets do this:

1.5*4 = 6

So she ran 6 kilometers in the first 4 days.

Next we have:

3*3 = 9

So she ran 9 kilometers in the last 3 days.

Now finally add them together:

6 + 9 = 15

So she ran a total of 15 kilometers.

Hope this helps!

7 0

3 years ago

Carlos is 33 years younger than Kristen. 2 years ago, Kristen's age was 2 times Carlos's age. How old is Carlos now?

ale4655 [162]

Answer:

Carlos is 35 years old now

Step-by-step explanation:

Assume that Carlos is x years old now

∵ Carlos is x years old now

∵ Carlos is 33 years younger than Kristen

- Kirsten's age is the sum of x and 33

∴ Kirsten is x + 33 years old now

2 years ago

∵ Carlos's age = x - 2

∵ Kristen's age = x + 33 - 2 = x + 31

∵ Kristen's age was 2 times Carlos's age

- Equate Kristen's age by 2 times Carlos's age

∴ x + 31 = 2(x - 2)

- Simplify the right hand side

∴ x + 31 = 2x - 4

- Subtract x from both sides

∴ 31 = x - 4

- Add 4 to both sides

∴ 35 = x

∵ x represents Carlos's age now

∴ Carlos is 35 years old now

6 0

3 years ago

What's another way to write 4 hundreds

liberstina [14]

400 is another way to right 4 hundreds

5 0

3 years ago

Read 2 more answers

Particle P moves along the y-axis so that its position at time t is given by y(t)=4t−23 for all times t. A second particle, part

sergey [27]

a) The limit of the position of particle $Q$ when time approaches 2 is $-\pi$ .

b) The velocity of particle $Q$ is $v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}}$ for all $t \ne 2$ .

c) The rate of change of the distance between particle $P$ and particle $Q$ at time $t = \frac{1}{2}$ is $\frac{4\sqrt{82}}{9}$ .

<h3>How to apply limits and derivatives to the study of particle motion</h3>

a) To determine the limit for $t = 2$ , we need to apply the following two algebraic substitutions:

$u = \pi t$ (1)

$k = 2\pi - u$ (2)

Then, the limit is written as follows:

$x(t) = \lim_{t \to 2} \frac{\sin \pi t}{2-t}$

$x(t) = \lim_{t \to 2} \frac{\pi\cdot \sin \pi t}{2\pi - \pi t}$

$x(u) = \lim_{u \to 2\pi} \frac{\pi\cdot \sin u}{2\pi - u}$

$x(k) = \lim_{k \to 0} \frac{\pi\cdot \sin (2\pi-k)}{k}$

$x(k) = -\pi\cdot \lim_{k \to 0} \frac{\sin k}{k}$

$x(k) = -\pi$

The limit of the position of particle $Q$ when time approaches 2 is $-\pi$ . $\blacksquare$

b) The function velocity of particle $Q$ is determined by the derivative formula for the division between two functions, that is:

$v_{Q}(t) = \frac{f'(t)\cdot g(t)-f(t)\cdot g'(t)}{g(t)^{2}}$ (3)

Where:

$f(t)$ - Function numerator.
$g(t)$ - Function denominator.
$f'(t)$ - First derivative of the function numerator.
$g'(x)$ - First derivative of the function denominator.

If we know that $f(t) = \sin \pi t$ , $g(t) = 2 - t$ , $f'(t) = \pi \cdot \cos \pi t$ and $g'(x) = -1$ , then the function velocity of the particle is:

$v_{Q}(t) = \frac{\pi \cdot \cos \pi t \cdot (2-t)-\sin \pi t}{(2-t)^{2}}$

$v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}}$

The velocity of particle $Q$ is $v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}}$ for all $t \ne 2$ . $\blacksquare$

c) The vector rate of change of the distance between particle P and particle Q ( $\dot r_{Q/P} (t)$ ) is equal to the vectorial difference between respective vectors velocity:

$\dot r_{Q/P}(t) = \vec v_{Q}(t) - \vec v_{P}(t)$ (4)

Where $\vec v_{P}(t)$ is the vector velocity of particle P.

If we know that $\vec v_{P}(t) = (0, 4)$ , $\vec v_{Q}(t) = \left(\frac{2\pi\cdot \cos \pi t - \pi\cdot t \cdot \cos \pi t + \sin \pi t}{(2-t)^{2}}, 0 \right)$ and $t = \frac{1}{2}$ , then the vector rate of change of the distance between the two particles:

$\dot r_{P/Q}(t) = \left(\frac{2\pi \cdot \cos \pi t - \pi\cdot t \cdot \cos \pi t + \sin \pi t}{(2-t)^{2}}, -4 \right)$

$\dot r_{Q/P}\left(\frac{1}{2} \right) = \left(\frac{2\pi\cdot \cos \frac{\pi}{2}-\frac{\pi}{2}\cdot \cos \frac{\pi}{2} +\sin \frac{\pi}{2}}{\frac{3}{2} ^{2}}, -4 \right)$

$\dot r_{Q/P} \left(\frac{1}{2} \right) = \left(\frac{4}{9}, -4 \right)$

The magnitude of the vector rate of change is determined by Pythagorean theorem:

$|\dot r_{Q/P}| = \sqrt{\left(\frac{4}{9} \right)^{2}+(-4)^{2}}$

$|\dot r_{Q/P}| = \frac{4\sqrt{82}}{9}$

The rate of change of the distance between particle $P$ and particle $Q$ at time $t = \frac{1}{2}$ is $\frac{4\sqrt{82}}{9}$ . $\blacksquare$

<h3>Remark</h3>

The statement is incomplete and poorly formatted. Correct form is shown below:

Particle $P$ moves along the y-axis so that its position at time $t$ is given by $y(t) = 4\cdot t - 23$ for all times $t$ . A second particle, $Q$ , moves along the x-axis so that its position at time $t$ is given by $x(t) = \frac{\sin \pi t}{2-t}$ for all times $t \ne 2$ .