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Misha Larkins [42]
3 years ago
10

Jacob has several babysitting jobs. On one job, he earned $10 an hour plus an extra $20 since there were 3 children. On the othe

r job, he earned $15 an hour
He earned the same amount of money at both jobs. How many hours did he work and how many hours did he earn?

Here are the equations
d= 15h
d= 10h+20

A: 2 hours,$40
B:4 hours $60
C: 3 hours $45
D: 5 hours $75
Mathematics
1 answer:
Slav-nsk [51]3 years ago
7 0

Answer:

4 hours, $60

Step-by-step explanation:

Assuming "d" is equal to the dollars Jacob earned and h= hours, you start by setting the equations equal to each other, since both are equal to "d."

15h=10h+20

Subtract 10h from both sides (to combine like terms) so the equation becomes:

5h=20

Divide 5 from both sides, to isolate "d" :

h=4

The number of hours Jacob worked is 4, as the equation above shows. To find the amount of money Jacob made, substitute 4 as "h" into the two original equations:

d= 15(4) d=$60

d=10(4)+20 d=$60

Jacob earned $60 dollars in 4 hours.

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For f(x) = 3x^2 – X+5, 2f (-3) - f(4)=
Alenkinab [10]

Answer:

3x^2-x+5.2f\left(-3\right)-f\left(4\right) =

3x^2-x-19.6f

Step-by-step explanation:

3x^2-x+5.2f\left(-3\right)-f\left(4\right)

=3x^2-x-5.2f\cdot \:3-f\cdot \:4

=3x^2-x-15.6f-4f

=3x^2-x-19.6f

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15-2=13
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Given the Arithmetic series A1+A2+A3+A4 7 + 11 + 15 + 19 + . . . + 91 What is the value of sum?
aivan3 [116]

Answer:

The value of the sum is 1078

Step-by-step explanation:

* Lets revise how to find the sum of the arithmetic series

- In the arithmetic series there is a constant difference between each

 two consecutive terms

- Ex:

# 4 , 9 , 14 , 19 , 24 , .......... (+ 5)

# 25 , 15 , 5 , -5 , -15 , .......... (-10)

- So if the first term is a and the constant difference between each two

  consecutive terms is d, then

  U1 = a , U2 = a + d , U3 = a + 2d , U4 = a + 3d , ..........

- Then the nth term is Un = a + (n - 1)d

- The sum of n terms is Sn = n/2[a + L] , where L is the last term in

 the series

* Lets solve the problem

∵ The arithmetic series is 7 + 11 + 15 + 19 + ......................... + 91

∴ The first term (a) is 7

∴ The last term is (L) 91

- Lets find the constant difference

∵ 11 - 7 = 4 and 15 - 11 = 4

∴ The constant difference (d) is 4

- Lets find the sum of the series from 7 to 91

∵ Sn = n/2[a + L]

∵ a = 7 and L = 91

∴ Sn = n/2[7 + 91]

- We need to know how many terms in the series

∵ L is the last term and equals 91 lets find its position in the series

∵ Un = a + (n - 1)d

∵ a = 7 , d = 4 and Un = 91

∴ 91 = 7 + (n - 1)(4) ⇒ subtract 7 from both sides

∴ 84 = (n - 1)(4) ⇒ divide both sides by 4

∴ 21 = n - 1 ⇒ add 1 to both sides

∴ n = 22

∴ The number of the terms in the series is 22

- Lets find the sum of the 22 terms (S22)

∴ S22 = 22/2[7 + 91]

∴S22 = 11[98] = 1078

* The value of the sum is 1078

4 0
3 years ago
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