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Jobisdone [24]
3 years ago
5

Write the equation of the line that has an undefined slope and passes through the point (4, 7). User: What is the slope of a lin

e that is parallel to the x-axis? User: What is the slope of the line that passes through the points (-3, 5) and (1, 7)? User: Which of the following equations is the translation 2 units left of the graph of y = |x|?
Physics
2 answers:
DerKrebs [107]3 years ago
6 0
<u>For the first question we use point-slope form of an equation</u>

y-y_{1}=m(x-x_{1})
y=m(x-x_{1})+y_{1}

then we plug in the known values

y=m(x-4)+7

we leave the slope as variable m since it is undefined

<u>For Question 2
</u>
A line that is parallel to the x-axis is a horizontal line and since the slope of a line is defined as the Δy/Δx (change in y/change in x) and the change in y is 0 at any 2 points observed on the line the slope is 0. (0/any number is 0)

<u>For Question 3
</u>
Following the same relationship as question 2 we can solve for the slope.

Δy/Δx
y_{1}-y _{2}/x_{1}-x_{2}

now we plug in the known values from the two points given

(5-7)/(-3-1)
-2/-4
m=1/2 or 0.5

<u><em>and For the Final Question</em></u>

a translation left or right is done by affecting the x variable if you add 2 to x then the x value will have to be 2 less to get the same result...in other words when x is 1 the value of y is also 1...but if I wanted the whole equation translated left 2 unites then I would want the same y-value at an x-value 2 smaller... in other words, in our example x will be -1 when y is 1. For this value found on the graph to match the equation our x value must have 2 added to it in the equation....therefore the equation that translates y=|x| two units left is...

y=|x+2| 



Brrunno [24]3 years ago
4 0

1.

Answer:

y = m(x - 4) + 7

Explanation:

as we know that equation of straight line passing through a fixed point and having undefined slope is given as

y - y_1 = m(x - x_1)

here we have

(x_1, y_1) = (4, 7)

so we will have

y - 7 = m(x - 4)

y = m(x - 4) + 7

2.

Answer:

slope = ZERO

Explanation:

Slope of the straight line is defined as the tangent of the angle made by the line with respect to x axis

here we need to find the slope of a straight line parallel to x axis so the angle is ZERO degree

hence the slope will be given as

m = tan 0

m = 0

3.

Answer:

m = 0.5

Explanation:

Slope of a straight line passing through two different points is given as

m = \frac{y_2 - y_1}{x_2 - x_1}

here we will have

m = \frac{7 - 5}{1 + 3}

m = \frac{2}{4}

m = 0.5

4.

Answer:

y = |x - x'|

here x' = any positive number

Explanation:

If we translate the graph 2 units left of the given position so we can say that we have shifted the graph above from its given position.

So we will have

y = |x - x'|

here x' = any positive number

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3 years ago
A cannon is fired from the edge of a cliff, which is 60m above the sea. The cannonball's initial velocity is 88.3m/s and it is f
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Answer:

a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m​

Explanation:

a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.

So y - y₀ = ut - 1/2gt²

y - y₀ = (v₀sinθ)t - 1/2gt²

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0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²

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So, 4.9t² - 50t - 60 = 0

Using the quadratic formula to find t,

t = \frac{-(-50) +/- \sqrt{(-50)^{2} - 4 X 4.9 X -60} }{2 X 4.9} \\t = \frac{50 +/- \sqrt{2500 + 1176} }{9.8} \\t = \frac{50 +/- \sqrt{3676} }{9.8} \\t = \frac{50 +/- 60.63 }{9.8} \\t = \frac{50 + 60.63 }{9.8} or t = \frac{50 - 60.63 }{9.8} \\t = \frac{110.63 }{9.8} or t = \frac{-10.63 }{9.8} \\t = 11.29 sor -1.085

Since t cannot be negative, t = 11.29 s

b. We first need to find the impact vertical velocity component. Using

v = u - gt where u = initial vertical velocity component = v₀sinθ  and t = 11.29 s and g = 9.8 m/s². So,

v = v₀sinθ - gt

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= √((3676 m²/s² + 5295.48 m²/s²)

= √(8971.48 m²/s²

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So the impact velocity is 94.72 m/s at -39.8°

c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m​

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