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Vladimir79 [104]
2 years ago
14

39+5h+4g-2h i have been trying 2 work this out 4 ages so plz help

Mathematics
2 answers:
SIZIF [17.4K]2 years ago
7 0
Unless you're given the value of the variables, you can't really solve for anything. The only thing you can do is simplify. 

<span>39+5h+4g-2h
The only like terms are 5h and -2h.

39 + 3h + 4g
This would be the simplified version.</span>
ryzh [129]2 years ago
7 0
Answer is 39 + 3h + 4y
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Using the percentile concept, it is found that 10 students scored lower than he did.

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When a measure is in the xth percentile of a data-set, it is greater than x% of the measures and lesser than (100 - x)%.

In this problem, he scored on the 20th percentile, that is, higher than 20% of the 50 students. Then:

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( C ) ΔMLN ≅ΔTSR

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Read 2 more answers
5. 3x + 5y + 5z =1<br> x - 2y = 5<br> 2x + 4y = 11
lilavasa [31]

Answer:

see explanation

Step-by-step explanation:

Given the 3 equations

3x + 5y + 5z = 1 → (1)

x - 2y = 5 → (2)

2x + 4y = 11 → (3)

Use (2) and (3) to solve for x and y

Multiply (2) by 2

2x - 4y = 10 → (4)

Add (3) and (4) term by term

4x = 21 ( divide both sides by 4 )

x = \frac{21}{4\\}

Substitute this value of x into (3)

2 × \frac{21}{4\\} + 4y = 11

\frac{21}{2\\} + 4y = 11 ( subtract \frac{21}{2\\} from both sides )

4y = \frac{1}{2} ( divide both sides by 4 )

y = \frac{1}{8\\}

Substitute the values of x and y into (1) and solve for z

3 × \frac{21}{4\\} + 5 × \frac{1}{8\\} + 5z = 1

\frac{63}{4} + \frac{5}{8} + 5z = 1

\frac{131}{8} + 5z = 1 ( subtract \frac{131}{8} from both sides )

5z = - \frac{123}{8} ( divide both sides by 5 )

z = - \frac{123}{40}

Solution is

x = \frac{21}{4\\}, y = \frac{1}{8\\}, z = - \frac{123}{40}

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