Luke ran 12.5 miles and 2 hours and 25 miles in 4 hours what is the constant of proportionality

What are the factors of 2x^2 + 5x -12?

Andrews [41]

Answer:

(2x - 3) • (x + 4)

Step-by-step explanation:

Step 1 :

Equation at step 1 :

(2x2 + 5x) - 12

Step 2 :

Trying to factor by splitting the middle term

2.1 Factoring 2x2+5x-12

The first term is, 2x2 its coefficient is 2 .

The middle term is, +5x its coefficient is 5 .

The last term, "the constant", is -12

Step-1 : Multiply the coefficient of the first term by the constant 2 • -12 = -24

Step-2 : Find two factors of -24 whose sum equals the coefficient of the middle term, which is 5 .

-24 + 1 = -23

-12 + 2 = -10

-8 + 3 = -5

-6 + 4 = -2

-4 + 6 = 2

-3 + 8 = 5

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -3 and 8

2x2 - 3x + 8x - 12

Step-4 : Add up the first 2 terms, pulling out like factors :

x • (2x-3)

Add up the last 2 terms, pulling out common factors :

4 • (2x-3)

Step-5 : Add up the four terms of step 4 :

(x+4) • (2x-3)

Which is the desired factorization

3 0

3 years ago

Help cant do it? It's trigonometry

12345 [234]

I think the answer to your problem is 69

6 0

3 years ago

A wire is to be attached to support a telephone pole. Because of surrounding buildings, sidewalks, and roadways, the wire must

ad-work [718]

This situation is a right triangle with the base measuring 18 and the hypotenuse 22. We are looking for x, the height up the pole that the wire is going to be attached to. We will use Pythagorean's Theorem to find the missing length. $22^2-18^2=x^2$ and $484-324=x^2$ . x^2 = 160 so x = $4 \sqrt{10}$ or 12.65 feet

3 0

3 years ago

Find the unit tangent vector T(t) to the curve r(t) = [sin(t), 1 + t, cos(t)] when t = 0.

ss7ja [257]

Compute the derivative of $\mathbf r(t)$ at $t=0$ - this will be the tangent vector - then normalize it by dividing it by its magnitude to get the unit tangent vector $\mathbf T(t)$ .

$\mathbf r(t) = \langle \sin(t), 1+t, \cos(t) \rangle \implies \dfrac{d\mathbf r}{dt} = \langle \cos(t), 1, -\sin(t) \rangle \implies \dfrac{d\mathbf r}{dt}(0) = \langle 1, 1, 0 \rangle$

$\|\langle1,1,0\rangle\| = \sqrt{1^2+1^2+0^2} = \sqrt2$

$\implies\mathbf T(0) = \dfrac{\langle1,1,0\rangle}{\sqrt2} = \boxed{\left\langle \dfrac1{\sqrt2}, \dfrac1{\sqrt2}, 0\right\rangle}$

7 0

1 year ago

Solve the quadratic equation by completing the square.<br> x^2+12x+52=0

White raven [17]

Answer:

(x + 6)² + 16 = 0

Step-by-step explanation:

To complete the square we will first need to get our equation to look like: x² + bx = c

Here we have x² + bx + c = 0 → x² + 12x + 52 = 0

First we need to subtract our c, in this case 52, from both sides to get x² + 12x = -52
We then need to add $(\frac{b}{2} )^{2}$ to both sides of the equation
Here our b value is 12, so plugging this into our formula we get $(\frac{12}{2} )^{2} =(6)^{2} =36$
Adding 36 to both sides our equation becomes: x² + 12x + 36 = -52 + 36
Then combining like terms on the right side we get x² + 12x + 36 = -16
Now making our left side of the equation into a perfect square we get: (x + 6)² = -16
Finally adding the 16 to both sides of the equation we get: (x + 6)² + 16 = 0

6 0

3 years ago