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mr_godi [17]
3 years ago
14

Solve x. 10(2x - 4) = 100x

Mathematics
1 answer:
mr_godi [17]3 years ago
3 0

Answer:

x=1/2

Step-by-step explanation:

Distribute the 10. it will be 20x-40=100x. Subtract 20x from both sides. it will be 40=80x. Divide by 80 to leave the x by itself. x=40/80...in simplify form....its 1/2

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Step-by-step explanation:

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How many solutions does the equation have?<br> x^4 + 7x^2 – 144 = 0
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You borrow $375 and pay back $20 each week what are the first 6 terms in the sequence
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Which operation results in an expression equivalent to ? 23y^4 - 6y^3 + 35y^2 - 20y
Tom [10]

Answer:

D. P + Q

Explanation:

Given:

P = 8y⁴ + 6y³ + 8y

Q = (5y² - 4y)(3y² + 7)

Required:

Determine which operation will give an expression equivalent to 23y⁴ - 6y³ + 35y² - 20y.

SOLUTION:

Perform each operation given to see which of them gives an expression equivalent to 23y⁴ - 6y³ + 35y² - 20y.

Q - P:

P = 8y⁴ + 6y³ + 8y

Q = (5y² - 4y)(3y² + 7)

Q - P = (5y² - 4y)(3y² + 7) - (8y⁴ + 6y³ + 8y)

(5y²(3y² + 7) -4y(3y² + 7)) - (8y⁴ + 6y³ + 8y)

(15y⁴ + 35y² - 12y³ - 28y) - (8y⁴ + 6y³ + 8y)

Open parentheses

15y⁴ + 35y² - 12y³ - 28y - 8y⁴ - 6y³ - 8y

Collect like terms

15y⁴ - 8y⁴ - 12y³ - 6y³ + 35y² - 28y - 8y

7y⁴ - 18y³ + 35y² - 36y

Therefore, P - Q does not give us an expression equivalent to 23y⁴ - 6y³ + 35y² - 20y.

PQ:

P = 8y⁴ + 6y³ + 8y

Q = (5y² - 4y)(3y² + 7) = (15y⁴ + 35y² - 12y³ - 28y)

PQ = (8y⁴ + 6y³ + 8y)(15y⁴ + 35y² - 12y³ - 28y)

=(8y⁴(15y⁴ + 35y² - 12y³ - 28y) +6y³(8y⁴ + 6y³ + 8y)(15y⁴ + 35y² - 12y³ - 28y) +8y(8y⁴ + 6y³ + 8y)(15y⁴ + 35y² - 12y³ - 28y))

= 120y⁸ . . . . .

Note: You don't need to perform this operation further anymore. It would definitely not give us the equivalent expression we are looking for. The degree of the leading term (120y⁸) is way too greater than the degree of the leading term (23y⁴) of the equivalent expression we are looking for.

Thus, PQ cannot give us an expression equivalent to 23y⁴ - 6y³ + 35y² - 20y.

P - Q:

P = 8y⁴ + 6y³ + 8y

Q = (5y² - 4y)(3y² + 7) = (15y⁴ + 35y² - 12y³ - 28y)

P - Q = (8y⁴ + 6y³ + 8y) - (15y⁴ + 35y² - 12y³ - 28y)

Open parentheses

8y⁴ + 6y³ + 8y - 15y⁴ - 35y² + 12y³ + 28y

Collect like terms

8y⁴ - 15y⁴ + 6y³ + 12y³ - 35y² + 8y + 28y

-7y⁴ + 18y³ - 35y² + 36y

Therefore, P - Q cannot give us an expression equivalent to 23y⁴ - 6y³ + 35y² - 20y.

P + Q:

P = 8y⁴ + 6y³ + 8y

Q = (5y² - 4y)(3y² + 7) = (15y⁴ + 35y² - 12y³ - 28y)

P + Q = (8y⁴ + 6y³ + 8y) + (15y⁴ + 35y² - 12y³ - 28y)

Open parentheses

8y⁴ + 6y³ + 8y + 15y⁴ + 35y² - 12y³ - 28y

Collect like terms

8y⁴ + 15y⁴ + 6y³ - 12y³ + 35y² + 8y - 28y

23y³ - 6y³ + 35y² - 20y

Therefore, P + Q will result in an expression equivalent to 23y⁴ - 6y³ + 35y² - 20y.

The answer is D.

7 0
3 years ago
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