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IgorLugansk [536]
3 years ago
14

if 7 dancers were at a dance and 46 more came in and h left and there was 13 dancers left how many was h

Mathematics
1 answer:
TEA [102]3 years ago
5 0
<span>If 7 dancers were at a dance and 46 more came in and h left and there was 13 dancers left how many was h
</span>-  h is 70.
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Multiple Choice
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A) Answer: A. 22.5 cm, 30 cm, and 37.5 cm


Setting the unity measurement as x, the three sides will be 3·x, 4·x, and 5·x. The sum of the three sides will be (3 + 4 + 5)·x = 12·x; this corresponds to the perimeter, therefore:

12·x = 90 cm


We can solve for x:

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Hence, the sides will be:

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7.5 · 4 = 30 cm

7.5 · 5 = 37.5 cm


This situation is represented in option A (which is also the only option whose sides add up to 90).



B) Answer: C. never true


We have:

6 + 8x - 9 = 11x + 14 - 3x


Bring all the terms with x on the left, by subtracting from both sides 11x and -3x, and all the numbers on the right by subtrcting from both sides 6 and -9:

6 + 8x - 9 - 11x - (-3x) - 6 - (-9) = 11x + 14 - 3x - 11x - (-3x) - 6 - (-9)


Cancel out the opposite terms from each side:

8x - 11x + 3x = 14 - 6 + 9


Combine likely terms:

0x = 17


Now you should ask yourself: what number multiplied by zero gives 17 as a result? The answer is none because a number multiplied by zero always equals zero.


Hence, the statement is never true.


C) Answer: r \leq \frac{1}{5}


We have:

5r + 4 ≤ 5


Subtract 4 from both sides:

5r + 4 - 4 ≤ 5 - 4


Combine likely terms:

5r ≤ 1


Divide both sides by 5:

\frac{5r}{5} \leq \frac{1}{5}

r \leq \frac{1}{5}


To graph the set of solutions, you need to set:

y₁ = 5x + 4 and

y₂ = +5


Plot these two lines:

y₂ is a horizontal line at a height of 5;

y₁ is a line passing through (0, 4) and (1, 9)


The set of solutions will be the part of the graph in which y₁ lies underneath y₂ (see picture attached), that happens for x \leq \frac{1}{5}.





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