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Elanso [62]
3 years ago
14

Sarah is training to run a race. She runs 1 mile per day for the first week. She doubles that and runs 2 miles each day for the

second week. She doubled her daily run again for the third week. How many miles did she run if she ran everyday those three week? Challenge question...
Mathematics
1 answer:
swat323 years ago
8 0

Answer:

She would have ran 48 miles, (if i did my math correctly)

Step-by-step explanation:

7 x 1 = 7

7 x 2 = 13

7 x 4 = 28

7+13+28 =48

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Help pls ASAP <br> SOLVING QUADRATIC EQUATIONS BY FACTORING
DIA [1.3K]

Answer:

Down, Down, Right, Right, diagonal left, up, right

Step-by-step explanation:

Not gonna do every equation here tho lol

7 0
2 years ago
Lee watches TV for 2 hours per day. During that time, the TV consumes 250 watts per hour. Electricity costs (10 cents)/(1 kilowa
vladimir1956 [14]

Answer:

$1.5

Step-by-step explanation:

Lets us first calculate energy consumed by TV in 30 days.

Energy consumed by TV in one hour : 250 watts.

TV operated 2 hours a day

so energy consumer by TV in a day = 250 * 2 watts = 500 watts hour

Energy consumed by TV in 30 day = energy consumer by TV in a day * 30

         = 500*30 watts = 15,000 watt-hour

Convert total energy consumed in kilo watt hour  

we know that one kilowatt hour = 1000 watt-hour

Thus, 15,000 watt-hour = 15,000/1000 kilowatt-hour = 15 kilowatt-hour

________________________________

cost of electricity = 10 cents per   kilowatt-hour

cost of electricity for 15 kilowatt-hour = 15* cost of electricity for one  kilowatt-hour( which is 10 cents)

=>15*10 = 150 cents

As $1 = 100 cents

hence 150 cents = $150/100 = $1.5 .

Thus, it cost $1.5 to to operate for a month of 30 days.

6 0
3 years ago
Which line is perpendicular to y=2/3x−5?<br> thanks!!
elixir [45]

Answer:

it's C

Step-by-step explanation:

did the quiz

8 0
3 years ago
A school track is shown the straightaway on each side measures 1,000 meters the curves are semicircles with diameter 74 meters w
lisov135 [29]

Answer:

2232.48 meters

Step-by-step explanation:

From the diagram:

Radius of semicircle = 74

Length of straightway on each side = 1000

The length of school track :

Straightway on each side = 2 * 1000 m = 2000m

Length of semicircle = πr/2

Length of both semicircle  = 2 * 3.142 * 74 /2 = 232.47785

Total Length = ( 2000 + 232.47785) = 2232.48 meters

7 0
2 years ago
The integral of (5x+8)/(x^2+3x+2) from 0 to 1
Gnom [1K]
Compute the definite integral:
 integral_0^1 (5 x + 8)/(x^2 + 3 x + 2) dx

Rewrite the integrand (5 x + 8)/(x^2 + 3 x + 2) as (5 (2 x + 3))/(2 (x^2 + 3 x + 2)) + 1/(2 (x^2 + 3 x + 2)):
 = integral_0^1 ((5 (2 x + 3))/(2 (x^2 + 3 x + 2)) + 1/(2 (x^2 + 3 x + 2))) dx

Integrate the sum term by term and factor out constants:
 = 5/2 integral_0^1 (2 x + 3)/(x^2 + 3 x + 2) dx + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx

For the integrand (2 x + 3)/(x^2 + 3 x + 2), substitute u = x^2 + 3 x + 2 and du = (2 x + 3) dx.
This gives a new lower bound u = 2 + 3 0 + 0^2 = 2 and upper bound u = 2 + 3 1 + 1^2 = 6: = 5/2 integral_2^6 1/u du + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx

Apply the fundamental theorem of calculus.
The antiderivative of 1/u is log(u): = (5 log(u))/2 right bracketing bar _2^6 + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx

Evaluate the antiderivative at the limits and subtract.
 (5 log(u))/2 right bracketing bar _2^6 = (5 log(6))/2 - (5 log(2))/2 = (5 log(3))/2: = (5 log(3))/2 + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx

For the integrand 1/(x^2 + 3 x + 2), complete the square:
 = (5 log(3))/2 + 1/2 integral_0^1 1/((x + 3/2)^2 - 1/4) dx

For the integrand 1/((x + 3/2)^2 - 1/4), substitute s = x + 3/2 and ds = dx.
This gives a new lower bound s = 3/2 + 0 = 3/2 and upper bound s = 3/2 + 1 = 5/2: = (5 log(3))/2 + 1/2 integral_(3/2)^(5/2) 1/(s^2 - 1/4) ds

Factor -1/4 from the denominator:
 = (5 log(3))/2 + 1/2 integral_(3/2)^(5/2) 4/(4 s^2 - 1) ds

Factor out constants:
 = (5 log(3))/2 + 2 integral_(3/2)^(5/2) 1/(4 s^2 - 1) ds

Factor -1 from the denominator:
 = (5 log(3))/2 - 2 integral_(3/2)^(5/2) 1/(1 - 4 s^2) ds

For the integrand 1/(1 - 4 s^2), substitute p = 2 s and dp = 2 ds.
This gives a new lower bound p = (2 3)/2 = 3 and upper bound p = (2 5)/2 = 5:
 = (5 log(3))/2 - integral_3^5 1/(1 - p^2) dp

Apply the fundamental theorem of calculus.
The antiderivative of 1/(1 - p^2) is tanh^(-1)(p):
 = (5 log(3))/2 + (-tanh^(-1)(p)) right bracketing bar _3^5


Evaluate the antiderivative at the limits and subtract. (-tanh^(-1)(p)) right bracketing bar _3^5 = (-tanh^(-1)(5)) - (-tanh^(-1)(3)) = tanh^(-1)(3) - tanh^(-1)(5):
 = (5 log(3))/2 + tanh^(-1)(3) - tanh^(-1)(5)

Which is equal to:

Answer:  = log(18)
5 0
3 years ago
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