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Ostrovityanka [42]
3 years ago
6

Find the least common factor denominator for these two rational 7/x^2 and 7/5x

Mathematics
1 answer:
MAVERICK [17]3 years ago
4 0
X^2 and 5x both have x in common to factor out. To add or subtract them, 5x^2 is the least common denominator .
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At a sale this week, a suit is being sold for $296.40. This is a 62% discount from the original price. What was the original pri
Soloha48 [4]
Suppose the original price to be x.
62%x=296.4
Divide both sides by 62% to get x=478, which is the original price.
7 0
2 years ago
Read 2 more answers
Solve and show all work |2×+3| -6=11​
mash [69]

Answer:

x = -10; x = 7  

Step-by-step explanation:

|2x + 3| - 6 =11

Add 6 to each side.

|2x + 3| = 17

Apply the absolute rule: If |x| = a, then x = a or x = -a.

(1) 2x + 3 = 17                    (2) 2x + 3 = -17

Subtract 3 from each side

         2x = 14                                 2x = -20

Divide each side by 2

          x = 7                                   x = -10

<em>Check: </em>

(1) |2(7) + 3| - 6 = 11                     (2) |2(-10) + 3| - 6 = 11

      |14 + 3| - 6 = 11                              |-20 + 3| - 6 = 11

             |17| - 6 = 11                                     |-17| - 6 = 1 1

              17 - 6 = 11                                        17 - 6 = 11

                    11 = 11                                              11 = 11

7 0
3 years ago
What ordered pair represents the number of jump ropes made per hour
V125BC [204]

where is the rest of the question?

is that it?

7 0
3 years ago
Read 2 more answers
Subtract<br> (2+8i)-(7+2i)
Mkey [24]
2 - 7 = -5
8i - 2i = 6i

-5 + 6i is your answer

hope this helps
3 0
3 years ago
Help please!<br><br> | 2x^2 + 5x + 3 | &gt; 0<br> solve the inequality
vovikov84 [41]

Factorize the quadratic.

2x^2 + 5x + 3 = (2x + 3) (x + 1)

We have |ab| = |a||b|, so

|2x^2 + 5x + 3| = |2x + 3| |x + 1| > 0

Now, both |2x+3|\ge0 and |x+1|\ge0 (since the absolute value of any number cannot be negative), so we just need to worry about when the left side is exactly zero. This happens for

(2x + 3) (x + 1) = 0 \implies 2x+3 = 0\text{ or }x+1 = 0 \\\\ \implies x = -\dfrac32 \text{ or } x = -1

So the solution to the inequality is the set

\left\{x \in \Bbb R \mid x\neq-\dfrac32 \text{ and } x\neq-1\right\}

3 0
2 years ago
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