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Viktor [21]
4 years ago
7

How do you solve this?

Mathematics
1 answer:
Murrr4er [49]4 years ago
8 0
(2x+5+3)(x+1) is the answer, you have to factor, hope that helps :)
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PLEASE SHOW YOUR WORK. thanks
astra-53 [7]

Answer:

75%

The drawings will help

Refer to the first one to help understand the second.

3 0
4 years ago
Eight friends decided to try white-water rafting on a nearby river. They rented four two-person rafts; one was yellow, one was g
stich3 [128]

Answer:

  1. Leroy Sands & Walt Smith - green
  2. Frank Cook & Phil Hawley - yellow
  3. Paul O'Brien & Alan Wilson - blue
  4. Henry Gladstone & Don Hughes - red

Step-by-step explanation:

First of all, it is convenient to choose the false statement.

We note that statements 7 (Alan is first) and 10 (Alan did not finish) are contradictory, so one of them must be false. Statement 5 has Henry & Hughes together, so is not consistent with statement 7 (Alan & Hughes). So, statement 7 is inconsistent with two other statements, thus must be the false one.

__

The attached table shows how we parsed the statements. An "O" signifies a combination that must occur; and "X" signifies one that cannot occur. We have used "-" to fill the spaces that are inconsistent with the locations of "O". The red letters are deduced from the others.

The first complete set we found was Henry & Hughes (Don & Gladstone) who were in the (red) capsized raft and did not finish. The second complete set we found was Phil & Cook (Frank & Hawley) who were in the (yellow) raft that finished second. We know Paul & Wilson (Alan & Partner) finished last, so Phil & Cook could not have been Paul and Wilson. Finally, O'Brien and Leroy were not on the same boat, so Paul must be Paul O'Brien and Leroy must be Leroy Sands.

4 0
3 years ago
You can use the what of a quadratic equation to determine the number and type of solutions of the equation?
adoni [48]

You can use the <u>determinant</u><u> (square root)</u> of a quadratic equation to determine the number and type of solutions of the equation.

\sqrt{b^{2}-4ac} > 0    → 2 real solutions

  • if \sqrt{b^{2}-4ac} is a perfect square, then rational sol'ns
  • if \sqrt{b^{2}-4ac} is not a perfect square, then irrational sol'ns

\sqrt{b^{2}-4ac} = 0    → 1 real solution (double root)

\sqrt{b^{2}-4ac} < 0    → 2 imaginary(complex) solutions

3 0
4 years ago
Can you guys help me with this question? PLEASEEEEE
Wittaler [7]

Answer:

90 dancers

Step-by-step explanation:

150/5=30

30x3=90

4 0
3 years ago
Read 2 more answers
Simplify (4 squared - 1+8÷8) ×5
DochEvi [55]
2-1+1*5= 6

just follow pemdas and simplify
5 0
4 years ago
Read 2 more answers
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