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galben [10]
3 years ago
6

George paid a total of $10,000 for stock that was $5 per share. If he sold all his shares for $15,000, how much profit on each s

hare did he make?
\$2000$2000
\$7.50$7.50
\$2.50$2.50
\$5$5
Mathematics
1 answer:
nika2105 [10]3 years ago
8 0

Answer:

OOGA BOOGA

Step-by-step explanation:

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Find the nature of the roots of the roots of the equation (x-705)(x-795)+800(x-750)(x-835)=0.
alexandr402 [8]

Answer:

B

Step-by-step explanation:

If you are allowed to use a graphing calculator, this can solve a lot of time. Graphing y = (x - 705)(x - 795) + 800(x - 750)(x - 835) gets us a parabola with x-intercepts of 749.97 and 834.924.

These roots are real and distinct, or choice B.

If you aren't allowed to use a graphing calculator, you can expand and combine like terms to get 801x^2 − 1269500x + 501560475. Looking at the discriminant, we see that (−1269500)^2 - 4*801*501560475 = 4630488100 is greater than 0 and not a perfect square, so it is real and distinct, or choice B.

I hope that helps!

5 0
2 years ago
Someone please help!!!!
aliya0001 [1]

Answer:

A

Step-by-step explanation:

You don't know whether they're all 90 degrees because we don't know about AB and CD.

7 0
3 years ago
Read 2 more answers
You cut square corners with side lengths that are whole numbers from a piece of cardboard with dimensions 20 inches by 30 inches
Orlov [11]
The correct answer is D 
8 0
3 years ago
How do I solve this equation?
Alenkinab [10]

Answer:

Step-by-step explanation:

This is an inequality, not an equation.  There will be a set of x-values for which the inequality will be true.

First eliminate the fractional coefficient 2/5 by multiplying all terms by 5:

25 < 2x + 15 ≤ 75

Subtracting 15 from each 5, (2/5)x, 3 and 15 yields:

22 < 2x ≤ 60

Dividing all three terms by 2, we get:

11 < x ≤ 30

Thus, any number greater than 11 but equal to or less than 30 is a solution.

7 0
3 years ago
What is the answer for this ??? please help
fredd [130]
It is 34 because 34*100 is 3400 so 3400/34=100

4 0
4 years ago
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