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mojhsa [17]
3 years ago
7

Solve for x x2 - 3|x - 2| - 4x = -6

Mathematics
1 answer:
kramer3 years ago
7 0

Answer:

x=2

Step-by-step explanation:

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Find the product.
dlinn [17]
The correct answer of the given expression above would be the last option. We just solved this using the distributive property.
So: <span>(a - 3)(a - 5)
a^2 -3a -5a + 15
a^2 -8a + 15
Hope this is the answer that you are looking for. Have a great day!</span>
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3 years ago
Eva has some money in her wallet. She spent $18.62 buying groceries and had $43.55 left. How much money did she have in her wall
jek_recluse [69]

Answer:

Step-by-step explanation:

62.17

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2 years ago
What is the solution to the following system of equations?
marshall27 [118]

Answer:

this is your answer

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3 years ago
Which numbers are irrational? Select all that apply.
ziro4ka [17]

Answer:

\sqrt{18} , \sqrt{78} , and π

Step-by-step explanation:

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\sqrt{18} , \sqrt{78} , and π

8 0
3 years ago
[50 pts] An even number, n, times the next consecutive number, n + 2, is equal to 168. Find the original number.
ale4655 [162]

<u>We are given:</u>

An even number 'n', multiplied by the next consecutive even number is 168

<u>Solving for n:</u>

From the given statement, we can say that:

n(n+2) = 168                      [<em>n multiplied by the next even number 'n+2'</em>]

n² + 2n = 168

n² + 2n - 168 = 0               [<em>subtracting 168 from both sides</em>]

We can see that we now have a quadratic equation, solving using splitting the middle term

n² + 14n - 12n - 168 = 0

n(n + 14) -12(n + 14) = 0      <em>[factoring out common terms</em>]

(n-12)(n+14) = 0

Here, we can divide both sides by either (n-12) OR (n+14)

Checking the result in both the cases:

(n + 14) = 0/(n-12)                     (n-12) = 0/(n+14)

n + 14 = 0                                 n - 12 = 0

n = -14                                      n = 12

Both these values are even and since we are not told if the number 'n' is positive or negative, both 12 and -14 are the possible values of n

4 0
3 years ago
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