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8090 [49]
2 years ago
12

What is the product? (-2d^2+s)(5d^2-6s)

Mathematics
1 answer:
wlad13 [49]2 years ago
5 0

The answer is −10d4+17d2s−6s2

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Verbal expression for 4x^5
dsp73
4 multiplied by x to the power of five
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3 years ago
Please help explain how to solve problem 16?
Anarel [89]
Here is the problem.....√(15x + 10) = 2x+3

to remove the square root, we do the opposite which is to square everything.

(√(15x + 10))² = (2x + 3)²      (the square negates the square root)
15x + 10 = (2x +3)(2x + 3)    (use the distributive property to continue)
15x + 10 = 4x² + 6x + 6x + 9  (combine like terms)
15x + 10 = 4x² + 12x + 9      (subtract 15x and 10 from each side)
-15x - 10          -15x - 10
0 = 4x² - 3x - 1                      (factor completely)
(x - 1) (4x + 1)                    (set each to equal 0)

x - 1 = 0                4x + 1 = 0
x = 1                      4x = -1
                               x = -1/4

place both into the equation to check for reasonableness...we see the negative number is not reasonable, but the x value of 1 is a solution.

answer is 1
4 0
3 years ago
Y is inversely proportional to (x + 2)
Ostrovityanka [42]

Answer:

y*(x+2)=k

so y= k/(x+2)

Step-by-step explanation:

3 0
3 years ago
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Help me plz and ty....​
nadya68 [22]

Answer: c

Step-by-step explanation:

8 0
3 years ago
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A right triangle has a hypotenuse length of 12, and one side length of 9. find the missing side length. do the side lengths form
Rina8888 [55]

Use the Pythagorean theorem:

a^2+b^2=c^2

c - a hypotenuse

a, b - legs

We have: a = 9 and c = 12. Substitute:

9^2+b^2=12^2

81+b^2=144               <em>subtract 81 from both sides</em>

b^2=63\to b=\sqrt{63}\\\\b=\sqrt{9\cdot7}\\\\b=\sqrt9\cdot\sqrt7\\\\ \boxed{b=3\sqrt{7}}

A Pythagorean triple consists of three positive integers a, b and c, such that

a^2+b^2=c^2.

b=3\sqrt{7} is not positive integer.

<h3>The sides of the triangle do not form a pythagorean triple.</h3>
5 0
3 years ago
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