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Annette [7]
3 years ago
10

Select the equation and the solution for the problem. let ( X ) represent the unknown number. a total of 48people in a sports te

am are riding 6 vans. how many people are riding in each van is carrying the same number of people.
(A) 48x = 6 ( x = 1/8
(B) 6x = 48 ( x = 8
(C) x + 6 = 48 ( x = 42)
(D) x - 6 = 48 ( x = 54)
Mathematics
2 answers:
devlian [24]3 years ago
7 0

Answer:

The answer is b because i did test and i want thoses points dawg

Step-by-step explanation:

Fantom [35]3 years ago
3 0

Answer:

B

Step-by-step explanation:

You might be interested in
A cable that weighs 8 lb/ft is used to lift 750 lb of coal up a mine shaft 500 ft deep. Find the work done. Show how to approxim
Annette [7]

Answer:

13.8 × 10⁵ lb/ft

Step-by-step explanation:

Suppose the distance(ft) below the top of the shaft is represented by x

The weight of the cable = 8 lb/ft

Weight of the coal to be lifted from mine = 750 lb

Recall that:

The work done by a force f to move an object through a distance x can be expressed as:

W = force (f) × displacement (x)

So, the force implies the total weight which should be lifted at any height x

f(x) = 750 + 8x

Using Riemann sum

where; the coal is lifted from x = 0 to x = 500 i.e. [a,b] = [0,500]

Dividing the interval into n subintervals

\Deltax = \dfrac{b -a }{n}

\Deltax = \dfrac{500- 0 }{n}

Suppose [x_{i-1},x_i ] to represent the i^{th} subinterval, then the work done can be estimated as:

W_i = f(x_i) \Delta x

W_i =(750 + 8x_i) \Delta x

Therefore; the total work done in between all the n subintervals is:

W = \sum \limits ^n_{i=1} W_1

W = \sum \limits ^n_{i=1} f(x_i) \Delta x

W = \sum \limits ^n_{i=1}(750 +8x_I)\Delta x

Therefore;

dW = f(x) dx

\int \ dW = \int \ f(x) \ dx where x ranges from 0 to 500

W = \int ^{500}_{0} \ 750 + 8x \ dx

W =\bigg  [750x + 4x^2 \bigg ]^{500}_{0}

W =\bigg  [750(500) + 4(500)^2-0 \bigg ]

W = 375000 + 1000000

W = 1375000

Thus; the total work done W = 13.8 × 10⁵ lb/ft

5 0
3 years ago
An investor bought 700 shares of stock, some at $2.25 per share and some at $2.75 per share. If the total cost was $1825.00, how
statuscvo [17]

The shares of $2.25 are 200 and the shares of $2.75 are 700.

According to the question,

We have the following information:

An investor bought 700 shares of stock, some at $2.25 per share and some at $2.75 per share. And the total cost was $1825.00.

Now, let's take the shares of $2.25 to be x and the shares of $2.75 to be y.

So, we have the following expression:

x+y = 700

x = 700-y ....(1)

2.25x+2.75y = 1825 ...(2)

Putting the value of x from equation 1:

2.25(700-y)+2.75y = 1825

1575-2.25y+2.75y = 1825

1575+0.5y = 1825

0.5y = 250

y = 250/0.5

y = 500

Now, putting the value of y in equation 1:

x = 700-y

x = 700-500

x = 200

Hence, the shares of $2.25 are 200 and the shares of $2.75 are 700.

To know more about shares here

brainly.com/question/21781662

#SPJ1

7 0
1 year ago
Hi please help ! this is rlly confusing! 15 points .
Bingel [31]
B. Quadrant 2 you can tell because point J is in the upper left quadrant.
4 0
3 years ago
Hi, I just have this one problem. Please explain how you got the answer as well, thanks!
Brums [2.3K]

2/3x = 6

Multiply both sides by 3x

2 = 6.3x

2 = 18x

x = 2/18

x = 1/9

7 0
3 years ago
Read 2 more answers
-m + 10 = -8 – 6m + 3m
Crank
M = -9 ; solve for m by simplifying both sides of the equation, then isolating the varible 
3 0
3 years ago
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