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dsp73
3 years ago
12

Luke sells toy cars for $12 each.His expenses are $3.50 per car,plus $34 for tools.How many cars must he sell for his revenue to

equal his expenses
Mathematics
1 answer:
faust18 [17]3 years ago
6 0

Answer: 4

Step-by-step explanation:

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HURRY PLEASE When 5 and 6 are multiplied by the same factor, how do the ratios compare to the ratio 5:6? The ratios are equivale
erma4kov [3.2K]

Answer:

The ratios are equivalent.

Step-by-step explanation:

If you multiply both 5 and 6, the ratio stays the same.

5:6 = 10:12 = 15:18 = 20:24 etc.

You can try this by dividing the larger number by the smaller number. You'll always get 1.2 as the "relationship", or ratio, is preserved.

24/20 = 1.2

15/18 = 1.2

10/12 = 1.2

6/5 = 1.2

6 0
3 years ago
Houlihan's age is two times Thomas's age. The sum of their ages is 57 . What is Thomas's age?
GarryVolchara [31]

year Thomas's is 17 :))))))))

4 0
4 years ago
Use the absolute value formula to express the distance between 9 and -8
Sonja [21]

Answer:

17

Step-by-step explanation:

| 9- (- 8) | = | 17 | = 17

Basically the absolute value of N1 - N2

3 0
3 years ago
Read 2 more answers
Is (a-3)(2a^2 + 3a + 3) equivalent to 2(a^3 - 1.5a^2 - 3a - 4.5)
vovikov84 [41]
<h3>Answer: Yes they are equivalent</h3>

==============================================

Work Shown:

Expand out the first expression to get

(a-3)(2a^2 + 3a + 3)

a(2a^2 + 3a + 3) - 3(2a^2 + 3a + 3)

2a^3 + 3a^2 + 3a - 6a^2 - 9a - 9

2a^3 + (3a^2-6a^2) + (3a-9a) - 9

2a^3 - 3a^2 - 6a - 9

Divide every term by 2 so we can pull out a 2 through the distributive property

2a^3 - 3a^2 - 6a - 9 = 2(a^3 - 1.5a^2 - 3a - 4.5)

This shows that (a-3)(2a^2 + 3a + 3) is equivalent to 2(a^3 - 1.5a^2 - 3a - 4.5)

4 0
3 years ago
Read 2 more answers
A bucket that weighs 5 lb and a rope of negligible weight are used to draw water from a well that is 60 ft deep. The bucket is f
mestny [16]

Answer:

The value is W= 2640 \  ft \cdot lb

Step-by-step explanation:

From the question we are told that

The weight of the bucket is F =  5 lb

The depth of the well is x_1 =  60 \ ft

The weight of the water is W_w  =  42 lb

The rate at which the bucket with water is pulled is v  = 1.5 \  ft/s

The rate of the leak is r = 0.15 lb/s

Generally the workdone is mathematically represented as

W =  \int\limits^{x_1}_{x_o} {G(x)} \, dx]

Here G(x) is a function defining the weight of the system (water and bucket ) and it is mathematically represented as

G(x) =  F  +  (W_w- Ix)

Here I is the rate of water loss in lb/ft mathematically represented as

I  = \frac{r}{v}

=> I  = \frac{0.15 }{1.5 }

=> I  = 0.1

So

G(x) =  5  +  (42- 0.1x)

=> G(x) =  47- 0.1x)

So

W =  \int\limits^{60}_{0} {47- 0.1x} \, dx]

=> W =  [47x - \frac{0.1x^2}{2} ]|\left 60} \atop {0}} \right.

=> W= [47(60) - 0.05(60)^2]

=> W= 2640 \  ft \cdot lb

7 0
3 years ago
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