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Radda [10]
3 years ago
9

camile mowed lawns for 3 hours and earned $6.80 per hour.then ahe wates windows for 2 hours and earns $6.20 per hour.what were c

amila's average earnings per hour for all 5 hours?
Mathematics
1 answer:
gulaghasi [49]3 years ago
3 0

Answer:

3 hours times 6.80= 20.40

2 hours times 6.20= 12.40

20.40 + 12.40 = 32.80

32.80 ÷ the 5 hours= $6.56 per hour average

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Sophia Thompson is 12 years old. Her mother is 5 years less than four times as old as Sophia. How old is mrs. Thompson? Please e
Ugo [173]
Sophia is 12. Four times that is 48. If her mother is 5 years less than this, she is 43.

(12 x 4) - 5 = 43
7 0
3 years ago
Which facts could be applied to simplify this expression? Check all that apply.
kykrilka [37]

The answer is B, C, and D. Like terms are terms with all the same variable, so 5x and -x are like terms.

C is correct. If we add -x to 5x, we get 4x. The other numbers remain unchanged because they have no like terms.

D is correct. Applying the rule of like terms, which is that like terms are numbers with the same variable, only add together numbers with the same variable.

Hope this helps!

3 0
3 years ago
Read 2 more answers
Prove 2√(x) + 1/√(x + 1) <= 2√(x+1) for all x in [0,inf)
EleoNora [17]
Start by multiplying each side of the inequality by \sqrt{x + 1} and simplifying:

2 \sqrt x + \frac{1}{\sqrt{x+1}}  \leq  2 \sqrt{x + 1}
(2 \sqrt x + \frac{1}{\sqrt{x+1}})(\sqrt{x + 1}) \leq (2 \sqrt{x + 1})(\sqrt{x + 1})
2 \sqrt{x(x + 1)} + 1 \leq 2(x + 1)
2 \sqrt{x^2 + x} + 1 \leq 2x + 2
2 \sqrt{x^2 + x} \leq 2x + 1
\sqrt{x^2 + x} \leq x + \frac{1}{2}

From here, we can square both sides to get

x^2 + x \leq (x + \frac{1}{2})^2
x^2 + x \leq x^2 + x + \frac{1}{4}
0  \leq \frac{1}{4}, which is always true.
3 0
3 years ago
Please help i have no idea what this is!!
Nesterboy [21]

Answer:

B

Step-by-step explanation:

y - y1 = m (x - x1)

y1 = 3

x1 = 5

m(the slope)= -12

7 0
3 years ago
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Please help if you dont mind?
lana [24]

Answer:

choice A

Step-by-step explanation:

7 0
3 years ago
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