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lyudmila [28]
3 years ago
5

Can someone help me please this thrue me off big time

Mathematics
2 answers:
Ronch [10]3 years ago
8 0
Y = mx + n

m slope = Δy / Δx 

for point(-7,43) and (4,-12)

Δy = -12 - 43 = -55
Δx = 4 - (-7) = 11

m = - 55/11 =  -5

choose a point for instance point (-7,43) and insert it into the equation, you get:

43 = -5 * (-7) + n = + 35 + n

n = 43 - 35 = 8

so: 

y = -5x + 8


Elena-2011 [213]3 years ago
4 0
Slope = (-12 - 43)/(4 +7) = -55/11 = -5

y = mx + b
43 = -5(-7) + b
43=35+b
b = 8

equation
y = -5x + 8
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Max is scuba diving at elevation of -64.5 feet, when friend signals to come higher. Max makes 2 ascents, each an equal distance
KengaRu [80]

Answer: - 42.95 feet

Explanation:

Let each ascent be x. Thus,

2 equal ascents = 2x

From the information given,

initial position = - 64.5 feet

Final position after 2 ascents = - 21.4 feet

This means that

- 64.5 + 2x = - 21.4

2x = - 21.4 + 64.5

2x = 43.1

x = 43.1/2

x = 21.55

Thus, Max's elevation after the first ascent is

- 64.5 + 21.55

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4 0
1 year ago
A lacrosse player throws a ball into the air from a height of 6 feet with an initial vertical velocity of 64 feet per second. Wh
-Dominant- [34]

Answer:

Step-by-step explanation:

I'm going to use calculus to solve this, because it's the simplest way.  

The acceleration due to gravity in feet is the second derivative of the position function.  We will start with the acceleration and work backwards with antiderivatives to get to the position function.

a(t) = -32.  Going backwards and using the fact that the initial vertical velocity is 64 ft/sec, our velocity function is

v(t) = -32t + 64.  Going backwards and using the fact that the initial height of the ball is 6 feet, our position function is

s(t)=-16t^2+64t+6

The first part of this question asks us the maximum height of the ball.  From Physics, we learn that the maximum height of a projectile is reached when the velocity is 0, which happens to be right where the projectile stops for a nanosecond in the air to turn around and come back down.  We set the velocity function equal to 0 and solve for t.

0 = -32t + 64 and

0 = -32(t - 2).  By the Zero Product Property, either -32 = 0 or t - 2 = 0.  It's obvious that -32 does not equal 0, so t - 2 must equal 0.  Solving this for t:

t - 2 = 0 so

t = 2 seconds.  Since the maximum height is reached at a time of 2 seconds, we plug 2 seconds into the position function to get its position at 2 seconds (which is also the max height of the ball).

s(2)=-16(2)^2+64(2)+6 and

s(2) = -64 + 128 + 6 so

s(2) = 70 feet

Now we want to know when the ball will hit the ground.  "When" is a time value, and we know that the height of the ball on the ground is 0, so we sub in a 0 for s(t) and factor the quadratic.

Using the quadratic formula:

t=\frac{-64+/-\sqrt{4096-4(-16)(6)} }{-32} and

t=\frac{-64+/-\sqrt{4480} }{-32} which gives us the 2 solutions

t=\frac{-64+\sqrt{4480} }{-32} and

t=\frac{-64-\sqrt{4480} }{-32}

Plugging into your calculator, the first t = -.0916500 and the second t = 4.091

We all know that time cannot ever be negative, so our t value is 4.09.

Again, from Physics, we know that a projectile reaches it max height at halfway through its travels, so it just goes to follow logically that if it halfway through its travels at 2 seconds, then it will hit the ground at 4 seconds.  And it does!! How awesome is that?!

4 0
3 years ago
A 1414​-ft ladder leans against a wall at a point 55 feet above the ground. how far is the bottom of the ladder from the​ wall?
viva [34]
Assume ladder length is 14 ft and that the top end of the ladder is 5 feet above the ground.

Find the distance the bottom of the ladder is from the base of the wall.

Picture a right triangle with hypotenuse 14 feet and that the side opposite the angle is h.  Then sin theta = h / 14, or theta = arcsin 5/14.  theta is

0.365 radian.  Then the dist. of the bot. of the lad. from the base of the wall is 

14cos theta = 14cos 0.365 rad = 13.08 feet.  This does not seem reasonable; the ladder would fall if it were already that close to the ground.

Ensure that y ou have copied this problem accurately from the original.

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Ulleksa [173]
I found this online, hope it helps


Just follow these steps:
Multiply normally, ignoring the decimal points.
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3 years ago
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