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3 years ago

What is the slope of

Mathematics

2 answers:

liraira [26]3 years ago

6 0

Answer:

3/2

Step-by-step explanation:

$y = mx + b$

where m is slope and b is y-intercept.

If we want to find the slope, you have to convert the equation into slope-intercept form. We have to move the whole equation to another side, leaving only y-term.

$2y - 3x = 7 \\ 2y = 7 + 3x \\ y = \frac{7 + 3x}{2}$

Simplify/Rearrange the equation into slope-intercept form.

$y = \frac{7}{2} + \frac{3}{2} x \\ y = \frac{3}{2} x + \frac{7}{2}$

Therefore, our slope is 3/2

earnstyle [38]3 years ago

4 0

Answer:

3/2 is slope

m= slope

y=mx+b

Step-by-step explanation:

2y=3/2x+7/2

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Pls help (mathematics)

olga2289 [7]

Answer:

I agree with all of them

Step-by-step explanation:

they are all connected mathematically (I used 3.14 I hope that the right idea)

7 0

2 years ago

A golfball is hit from the ground with an initial velocity of 200 ft/sec. The horizontal distance that the golfball will travel,

algol13

Answer:

The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

Step-by-step explanation:

The formula from the maximum distance of a projectile with initial height h=0, is:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}$

Where $v_i$ is the initial velocity.

In the closed interval method, the first step is to find the values of the function in the critical points in the interval which is $[0, \pi/2]$ . The critical points of the function are those who make $d'(\theta)=0$ :

$d(\theta)=\frac{v_i^2\sin(2\theta)}{g}\\d'(\theta)=\frac{v_i^2\cos(2\theta)}{g}*(2)\\d'(\theta)=\frac{2v_i^2\cos(2\theta)}{g}$

$d'(\theta)=0\\\frac{2v_i^2\cos(2\theta)}{g}=0\\\cos(2\theta)=0\\2\theta=\pi/2,3\pi/2,5\pi/2,...\\\theta=\pi/4,3\pi/4,5\pi/4,...$

The critical value inside the interval is $\pi/4$ .

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\d(\pi/4)=\frac{v_i^2sin(2(\pi/4))}{g}\\d(\pi/4)=\frac{v_i^2sin(\pi/2)}{g}\\d(\pi/4)=\frac{v_i^2(1)}{g}\\d(\pi/4)=\frac{(200)^2}{32}\\d(\pi/4)=\frac{40000}{32}\\d(\pi/4)=1250ft$

The second step is to find the values of the function at the endpoints of the interval:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\\theta=0\\d(0)=\frac{v_i^2sin(2(0))}{g}\\d(0)=\frac{v_i^2(0)}{g}=0ft\\\theta=\pi/2\\d(\pi/2)=\frac{v_i^2sin(2(\pi/2))}{g}\\d(\pi/2)=\frac{v_i^2sin(\pi)}{g}\\d(\pi/2)=\frac{v_i^2(0)}{g}=0ft$

The biggest value of f is gived by $\pi/4$ , therefore $\pi/4$ is the absolute maximum.

In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

4 0

3 years ago

8-6: MathXL for School: Practice & Problem Solving

notka56 [123]

Answer:

The Area of a circle is

$\pi {r}^{2}$