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Sloan [31]
3 years ago
11

5. Grace has $39 She has 8 less than her brother. How much money does Grace's brother have?

Mathematics
2 answers:
natta225 [31]3 years ago
8 0
He has $47. If you need me to explain then I will
alukav5142 [94]3 years ago
6 0

Answer:

Grace's brother has $47.

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A study on the population of wolves is being done in a county to determine how the number of wolves has changed over time. The p
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Answer:

<h2>The year is 2010.</h2>

Step-by-step explanation:

The population of wolves is given by P(t) = 80(0.98)t.

The study has began in the year of 2000 and will end in 2020.

The difference between 2000 and 2009 is 9 years and the difference between 2009 and 2020 is 11 years.

Hence, the maximum value of t is 11.

The domain is 0 \leq t \leq 11.

The population will be 80, that is P(t) = 80.

Hence, 0.98t = 1, or, t = \frac{1}{0.98} = \frac{100}{98}  = \frac{50}{49} ≅1.

In (2009 + 1) = 2010, the population of wolves will be 80.

3 0
3 years ago
the area of a square whose sides have length x cm is increasing at the rate of 12cm²/s. how fast is the length of a side increas
katovenus [111]

Answer:

0.67 cm/s

Step-by-step explanation:

The area of a square is given by :

A=x^2 ....(1)

Where

x is the side of a square

\dfrac{dA}{dt}=12\ cm^2/s

Differentiating equation (1) wrt t.

\dfrac{dA}{dt}=2x\times \dfrac{dx}{dt}

When A = 81cm²​, the side of the square, x = 9 cm

Put all the values,

12=2\times 9\times \dfrac{dx}{dt}\\\\\dfrac{dx}{dt}=\dfrac{2}{3}\\\\\dfrac{dx}{dt}=0.67\ cm/s

So, the length of the side of a square is changing at the rate of 0.67 cm/s.

4 0
2 years ago
Consider the given geometric sequence.
Gala2k [10]
F5=(256)f1=16 hope this helps
7 0
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The sum of a rational number and irrational number is irrational.
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Answer:

A) Always True

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The distance from Charles’s house to his school is 3 miles. His father gave him a ride for 1,760 yards, and he walked the rest o
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The answer is 2 miles or 3520 yards. (they are the same answer.)

If this helps, can you give me the brainliest answer? I need it...

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