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2 years ago

8

Write an equation of the line in either point slope or slope intercept form that passes through the points (3,5) and (2,-2). (Hi

nt: Calculate the slope. )

1 answer:

Jobisdone [24]2 years ago

6 0

Answer:

Step-by-step explanation:

slope= y2-y1 / x2-x1

= -2-5 / 2-3

= -7 / -1

= 1

You might be interested in

Part A: A graph passes through the points (0, 4), (1, 8), and (2, 10). Does this graph represent a linear function or a non-line

IRISSAK [1]

Part A: In (0,4) and (1,8), as x increases from 0 to 1, y increases from 4 to 8. The slope of the line connecting these 2 pts. is (8-4) / (1-0), or 4. Now do the same thing for the 2 pts (1,8) and (2, 10). You will find that the slope in the 2nd case is different. Thus, the 3 given pts do NOT lie on a straight line. Function is non-linear.

Part B: an example of a non-linear function would be y = 2x^2 - 5. You can tell immediately from that exponent, 2, that this function is non linear.

An ex. of a linear function would include x to the 1st power only: y = 3x - 7.

4 0

3 years ago

Read 2 more answers

PLS HELP ME I WILL MARK YOU AS BRAINLIEST!!!

Scilla [17]

Answer:

volume = length x width x height

v=11x5x11 = 605

hope that answers your question

8 0

2 years ago

Help me with this please !!

avanturin [10]

Answer:

see explanation

Step-by-step explanation:

Under a reflection in the line y = - x

a point (x, y ) → (- y, - x ), thus

T(- 1, 3 ) → T'(- 3, 1 )

U(- 1, 10 ) → U'(- 10, 1 )

V(- 2, 4 ) → V'(- 4, 2 )

6 0

3 years ago

3(2x+1)=7-2(x+5)<br> Can you please show me how to solve this step by step please

Talja [164]

3(2x+1) = 7-2(x+5)
6x+3 = 7-2x-10
-3 -3
6x = 7-2x-10-3
+2x +2x
8x = -6
---- -----
8 8
x = -3/4

8 0

3 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