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Elenna [48]
3 years ago
13

Pls someone answer quickly!!!!!

Mathematics
1 answer:
Daniel [21]3 years ago
8 0
Yes, EF will just be 79.
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Lonny is a salesman and earns $75 for each sale he makes. Lonny has a sales
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5x-7+8=-27

Step-by-step explanation:

5x-7+8=-27

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What is 8 times 6 times 10 equals
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Step-by-step explanation:

8x6x10=480

48x10

480

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Suppose you own two stocks, A and B. In year 1, stock A earns a 2% return and stock B earns a 9% return. In year 2, stock A earn
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Answer:

Both the stock have the same expected return.

Step-by-step explanation:

In year 1 the return earned by stocks A and B are:

Stock A = 2% return

Stock B = 9% return

In year 2 the return earned by stocks A and B are:

Stock A = 18% return

Stock B = 11% return

Compute the expected return for stock A as follows:

Expected\ return\ for\ A=\frac{2+18}{2}=10\%

Compute the expected return for stock B as follows:

Expected\ return\ for\ B=\frac{9+11}{2}=10\%

Thus, both the stock have the same expected return.

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3 years ago
Show two different ways to factor −4x −28.
vazorg [7]

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-4x - 28 = -4(x + 7) or 4(-x - 7)

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3 years ago
A basketball player, standing near the basket to grab a rebound, jumps 66.1 cm vertically. How much time does the player spend i
Zepler [3.9K]

Answer:

0.293 s

Step-by-step explanation:

Using equations of motion,

y = 66.1 cm = 0.661 m

v = final velocity at maximum height = 0 m/s

g = - 9.8 m/s²

t = ?

u = initial takeoff velocity from the ground = ?

First of, we calculate the initial velocity

v² = u² + 2gy

0² = u² - 2(9.8)(0.661)

u² = 12.9556

u = 3.60 m/s

Then we can calculate the two time periods at which the basketball player reaches ths height that corresponds with the top 10.5 cm of his jump.

The top 10.5 cm of his journey starts from (66.1 - 10.5) = 55.6 cm = 0.556 m

y = 0.556 m

u = 3.60 m/s

g = - 9.8 m/s²

t = ?

y = ut + (1/2)gt²

0.556 = 3.6t - 4.9t²

4.9t² - 3.6t + 0.556 = 0

Solving the quadratic equation

t = 0.514 s or 0.221 s

So, the two time periods that the basketball player reaches the height that corresponds to the top 10.5 cm of his jump are

0.221 s, on his way to maximum height and

0.514 s, on his way back down (counting t = 0 s from when the basketball player leaves the ground).

Time spent in the upper 10.5 cm of the jump = 0.514 - 0.221 = 0.293 s.

5 0
3 years ago
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