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swat32
3 years ago
6

A baseball pitching machine stands about 2 feet tall and throws a baseball upward at a velocity of 50 feet per second.

Mathematics
1 answer:
horrorfan [7]3 years ago
7 0

so what is the question?

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Find the slope of the line that passes through the pair of points. <br> F(–6, 8), P(–6, –5)
asambeis [7]

-5-8 ÷ -6--6

-13 ÷ 6

gradient = -13/6

6 0
4 years ago
Choose the algebraic equation that matches the table.
dexar [7]

the correct answer is A. Y=X+2
3 0
3 years ago
A car is manufactured for a cost of $12,500. The dealership purchases the car and puts a markup of 67% on the car. The car is pu
Naya [18.7K]
<h3>Answer:</h3>
  1. $12,500 — dealer cost
  2. $20,875 — owner cost
  3. $5385.75 — owner's selling price
<h3>Step-by-step explanation:</h3>

The problem statement is missing a lot of information, so we'll make some assumptions:

  • the cost of manufacture is the dealer's cost
  • the markup percentage uses the dealer's cost as a base
  • the car is purchased for the marked price in each case.

1. The dealer's cost is the cost of manufacture: $12,500.

2. The marked-up price is

  $12,500 + 0.67×12,500 = $20,875.

3. The book value after 8 years is

  $20,875 - 0.57×20,875 = $8976.25

So, the Craigslist price is

  0.60×$8976.25 = $5385.75

4 0
3 years ago
Solve the following equation for x: z=x^2+wx
Alja [10]
Minus z both sides
0=x^2+wx+z
we can use quadratic formula or manually complete the square
let's use quadratic

for
0=ax^2+bx+c
x= \frac{-b+/- \sqrt{b^2-4ac} }{2a}
given
0=1x^2+wx-z
a=1
b=w
c=-z

x= \frac{-w+/- \sqrt{w^2-4(1)(-z)} }{2(1)}
x= \frac{-w+/- \sqrt{w^2+4z} }{2}

x= \frac{-w+ \sqrt{w^2+4z} }{2} or x= \frac{-w- \sqrt{w^2+4z} }{2}
4 0
3 years ago
A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters
In-s [12.5K]

Answer:

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

Step-by-step explanation:

Given that,

A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

\Rightarrow y=\sqrt{x^2+100}

Differentiating with respect to t

\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}

\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}

Since the car driving towards the intersection at 13 m/s.

so,\frac{dx}{dt}=-13

\therefore \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}.(-13)

Now

\therefore \frac{dy}{dt}|_{x=24}=\frac{24}{\sqrt{24^2+100}}.(-13)

               =\frac{24\times (-13)}{\sqrt{676}}

               =\frac{24\times (-13)}{26}

               = -12 m/s

Negative sign denotes the distance between the car and the person decrease.

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

8 0
3 years ago
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