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3 years ago

Three hundred cars drove over a bridge in 23 minutes. At that rate, how

Mathematics

2 answers:

anzhelika [568]3 years ago

7 0

Answer:

1800

Step-by-step explanation:

300 cars in 23 minutes so 300/25 = how many cars each minute

that equals 13.04347....

so on your calculator multiply that by 138 which gives you 1800 cars

Paraphin [41]3 years ago

3 0

Answer:

138/23×300=1800cars drove over the bridge

Step-by-step explanation:

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4x ≥ 12 What does it mean?

Pepsi [2]

Answer:

X = 3 or higher

Step-by-step explanation:

hope it helps, let me know if it doesn't :)

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3 years ago

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The sum of three consecutive numbers is -9 what are the numbers

zloy xaker [14]

-9/ 3= -3

-3 + -4 + -2 = -9

the numbers are -2 , -3 , -4

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3 years ago

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Mari spends $6 on cheese for a party she also plans to buy b boxes of crackers that cost $3 per box

PSYCHO15rus [73]

Answer:

The answer is 6+3b<or equal to 15

Step-by-step explanation:

3 0

3 years ago

if you took 7 hours to mow 4 lawns, then at that rate, how many lawns could be lawns could be mowed in 35 hours? at what rate we

liraira [26]

It takes 7 hours to mow 4 lawns
Therefore every hour, (4 ÷ 7) or 4/7 lawns are mowed. This gives the rate.

For 35 hours, 4/7 x 35 lawns are mowed.
=35 ÷ 7 = 5
= 4 × 5
= 20 lawns

3 0

3 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

2 years ago