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uranmaximum [27]
3 years ago
9

Helppppppppppppp pleaseeeeeeeeeeeeeeee

Mathematics
1 answer:
strojnjashka [21]3 years ago
4 0

Answer:

The answer would be b, 8:13

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Melody can cycle for 30 miles against the wind in the same amount of
zimovet [89]

Answer:

15x/2*4

Step-by-step explanation:

6 0
2 years ago
In the expression 5 n minus StartFraction 2 m over 7 EndFraction + three-fourths, what is the constant?
SIZIF [17.4K]

Answer:

<u>The correct answer is B. 3/4</u>

Step-by-step explanation:

In the expression 5n - 2m/7+ 3/4, what is the constant?

Let's analyze the coefficients, variables, operators and constants, as follows:

A. Coefficients:

5 is coefficient for n and 2/7 is coefficient for m

B. Variables:

n and m

C. Operators:

- between  5n - 2m/7

+ between 2m/7+ 3/4

D. Constant:

3/4 that is not affected by any variable

<u>The correct answer is B. 3/4</u>

7 0
3 years ago
Read 2 more answers
Please I need help I been stuck on this problem for 40 minutes
nignag [31]
First
224-108=116miles left
And 4 days left
116/4
=29 miles each day
5 0
3 years ago
Read 2 more answers
The rate of residential electricity consumption​ (in billions of kilowatt hours​ (kWh) per​ year) of a country is approximated b
Fiesta28 [93]
Congrats on making it to integrals!
Basically you need to integral your function because integral rate in respect to time = total amount.
Also your bounts are (2001-1990,2006-1990)=(11,16)

Thus we take integral like:
int(11,16)(928.5e^(0.0249x))=
(11,16)928/0.0249e^(0.0249(x))-928/0.0249e^(0.0249(x))
(You can check this by taking its derivative and seeing if you get the original function)
928/0.0249e^(0.0249(16))-928/0.0249e^(0.0249(11))=6498.1
4 0
3 years ago
The number of gallons of water in a large tank at time (t) minutes is given by the function W(t)=160,000-8,000t+t². Find the ave
sergij07 [2.7K]

The number of gallons of water in the tank at t=10 is

... W(10) = 160,000 -10(8000 -10) = 80100

The number of gallons of water in the tank at t=10.5 is

... W(10.5) = 160,000 -10.5(8000 -10.5) = 76110.25

The rate of change over the interval is

... (W(10.5) - W(10))/(10.5 - 10) = (76110.25 - 80100)/(0.5) = -7979.5

The average rate of change in the number of gallons of water in the tank over the interval is -7979.5 gal/min.

The sign is negative, so the amount of water is decreasing.

3 0
3 years ago
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