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2 years ago

What is the slope for y = 2x + 5

2 answers:

Crank2 years ago

4 0

Answer: 2

(The slope is always the coefficient of the independent variable, in this case x . The slope is therefore 2.)

Darya [45]2 years ago

4 0

Answer:2

Step-by-step explanation:

I use math-way and it helps a lot

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What is the standard deviation of a normal distribution, whose mean is 35, in which an x-value of 23 has a z-score of -1.63?

Dvinal [7]

Answer:

7.36

Step-by-step explanation:

z-score is:

z = (x − μ) / σ

Given that z = -1.63, x = 23, and μ = 35:

-1.63 = (23 − 35) / σ

σ ≈ 7.36

6 0

3 years ago

Pipe A can fill a swimming pool in 12 h. Working with another Pipe B, it only takes 3h How long would it take Pipe B working al

MatroZZZ [7]

My guess is 9 hours, because since pipe a takes 12 hours and with pipe b it takes 9, it's 12-9=3 I think.

5 0

3 years ago

The path of a volley ball thrown over a net is modeled with the function A(x) = -0.02x2 + 0.6x + 5, where x is

LiRa [457]

Answer:

The ball travel horizontally 37 feet before it hits the ground

Step-by-step explanation:

we have

$A(x)=-0.02x^2+0.6x+5$

where

x ----> is the horizontal distance in feet

A ---> is the altitude of the ball, in feet.

we know that

When the ball hit the ground, the altitude of the ball is equal to zero

For A(x)=0

$-0.02x^2+0.6x+5=0$

solve the quadratic equation

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-0.02x^2+0.6x+5=0$

$a=-0.02\\b=0.6\\c=5$

substitute in the formula

$x=\frac{-0.6\pm\sqrt{0.6^{2}-4(-0.02)(5)}} {2(-0.02)}$

$x=\frac{-0.6\pm\sqrt{0.76}} {-0.04}$

$x=\frac{-0.6\pm0.87} {-0.04}$

$x=\frac{-0.6+0.87} {-0.04}=-6.75$

$x=\frac{-0.6-0.87} {-0.04}=36.75$

therefore

The ball travel horizontally 37 feet before it hits the ground

3 0

3 years ago

If x = a cosθ and y = b sinθ , find second derivative

Olin [163]

I'm guessing the second derivative is for <em>y</em> with respect to <em>x</em>, i.e.

$\dfrac{\mathrm d^2y}{\mathrm dx^2}$

Compute the first derivative. By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta$

$x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta$

Now compute the second derivative. Notice that $\frac{\mathrm dy}{\mathrm dx}$ is a function of $\theta$ ; so denote it by $f(\theta)$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}$

By the chain rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$