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sasho [114]
3 years ago
15

A shipment to a warehouse consists of 4 large boxes weighing 18.5 pounds each, 2 medium boxes weighing 6.75 pounds each, and 14

small boxes weighing 78 pounds each.
What is the total weight, in pounds, of this shipment?
Mathematics
1 answer:
vagabundo [1.1K]3 years ago
8 0

Answer:

99.75lbs.

Step-by-step explanation:

4 x 18.5=    74lbs

2  x6.75= 13.5

14x 7/8 =98/8=12.25

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5.) -6x – 29 = 5x – 7
Umnica [9.8K]

Answer: x= -2

Step-by-step explanation: first step is to add 7 to both sides. You’re left with -6x-22=5x. After this you’re going to add 6 to both sides. You get -22=11x. Now you divide both sides by 11 to get X by itself. You’re left with -2.

6 0
3 years ago
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Translate the sentence into an equation.
Dmitriy789 [7]

Answer:

I think it could be 8+6+y

Step-by-step explanation: I dont really know but dont delete this. Can I be Brainlyest please?

5 0
2 years ago
One gallon of gasoline in buffalo, New York costs $2.29. In Toronto, Canada, one liter of gas costs $0.91. There are 3.8 liters
Solnce55 [7]

<u>Answer:</u>

The cost of one gallon of gas in Toronto, Canada is $3.458.

<u>Solution: </u>

Given, One gallon of gasoline in buffalo, New York costs $2.29.  

In Toronto, Canada, one liter of gas costs $0.91.  

There are 3.8 liters in one gallon.  

We have to find how much does one gallon of gas cost in Toronto?

Now, cost of one gallon of gas in Toronto = cost of one litre of gas in Toronto x 3.8 liters for 1 gallon.

Cost of one gallon of gas = 0.91 x 3.8 = 3.458

Hence, the cost of one gallon of gas in Toronto, Canada is $3.458.

3 0
3 years ago
Consider the function f given by f(x)=x*(e^(-x^2)) for all real numbers x.
NISA [10]

Answer:

\frac{\sqrt{\pi}}{4}

Step-by-step explanation:

You are going to integrate the following function:

g(x)=x*f(x)=x*xe^{-x^2}=x^2e^{-x^2}  (1)

furthermore, you know that:

\int_0^{\infty}e^{-x^2}=\frac{\sqrt{\pi}}{2}

lets call to this integral, the integral Io.

for a general form of I you have In:

I_n=\int_0^{\infty}x^ne^{-ax^2}dx

furthermore you use the fact that:

I_n=-\frac{\partial I_{n-2}}{\partial a}

by using this last expression in an iterative way you obtain the following:

\int_0^{\infty}x^{2s}e^{-ax^2}dx=\frac{(2s-1)!!}{2^{s+1}a^s}\sqrt{\frac{\pi}{a}} (2)

with n=2s a even number

for s=1 you have n=2, that is, the function g(x). By using the equation (2) (with a = 1) you finally obtain:

\int_0^{\infty}x^2e^{-x^2}dx=\frac{(2(1)-1)!}{2^{1+1}(1^1)}\sqrt{\pi}=\frac{\sqrt{\pi}}{4}

5 0
3 years ago
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Although the actual amount varies by season and time of day the average volume of water that falls over the falls each second is
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262.8*10 to the 7th power
3 0
3 years ago
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