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lana [24]
3 years ago
12

Isaiah already has 16 dollars. Every hour (x) that he works at the bakery, he earns another 4 dollars (y). How many total dollar

s will Isaiah have after 13 hours of work?
Mathematics
2 answers:
iVinArrow [24]3 years ago
7 0
Its 68
He already has $16
If he makes $4 an hour and does 13 hours of work then he would make
$52
$52+$16=$68
Therefore he will have $68 in total
Lena [83]3 years ago
5 0
$68
4x13=52 + 16 = 68
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\boxed{(2+y)^3-cube\ the\ sum\ of\ the\ numbers\ 2\ and\ y}

Use:\ (a+b)^3=(a+b)(a+b)(a+b)=a^3+3a^2b+3ab^2+b^3\\----------------------------\\(2+y)^3=2^3+3\cdot2^2\cdot y+3\cdot2\cdot y^2+y^3=8+3\cdot4y+6y^2+y^3\\\\=8+12y+6y^2+y^3\\========================================

8+12y+6y^2+6y^3+8y+4y^2+y^3\\=8+(12y+8y)+(6y^2+4y^2)+(6y^3+y^3)\\=8+20y+10y^2+7y^3



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3 years ago
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How do you do 5+5?
steposvetlana [31]

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If you had 5 apples and Johnny gave you five more how much do you have? Lets count

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6 0
3 years ago
This purse is roughly a trapezoid. The bottom base is 16 inches long. The top base is 10. It stands 13 inches tall. The latch is
balandron [24]
To find the amount of fabric for one side of the purse you will find the area of the trapezoidal space that is created.  The latch is on the top, so the 2 in is not necessary information.

A = 1/2 h(b1 + b2)
      1/2 x 13 x (10 + 16)
A = 169 square inches of fabric will be needed.
3 0
2 years ago
Helen had $12 but spent $3 of it. What per cent did she spend?
Scrat [10]

Answer:

25%

Step-by-step explanation:

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6 0
1 year ago
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Some of the students three scores is 231. If the first is 20 points more than the second, and the sum of the first two is 6 more
Aleks [24]

Answer:

The first score is 109

Step-by-step explanation:

I am assuming that in the first sentence of the question, you meant:

Sum of the students three scores is 231...

First, let the scores of the first second and third student be a, b and c respectively. We are told that:

a + b + c = 231 . . . . . . . .(1)         (sum of students three scores is 231)

a = b + 20 . . . . . . . . . . . (2)        (the first is 20 points more than the second)

a + b = 6c . . . . . . . . . . . .(3)        (sum of the first two is 6 more times the third)

required, find a.

substituting the value of (a + b) in equation (3) into equation (1), we will have the following:

since a + b = 6c . . . (3)

a + b + c = 231 . . . . . (1), becomes,

(a + b) + c = 231

(6c)  + c = 231

7c = 231 (divide both sides by 7)

c = 231 ÷ 7 = 33

∴ c = 33

Next, from equation (2), we know that a = b + 20; this can also be written as:

a - 20 = b

∴ b = a - 20 . . . . . . . (4)

Finally, putting the value of b in equation (4) and the value of c calculated above into equation 1, ( a + b + c = 231), we have the following:

a + (a - 20) + 33 = 231

(a + a) - 20 + 33 = 231

2a + 13 = 231

2a = 231 - 13 = 218

a = 218 ÷ 2 = 109

∴ a = 109

we can also calculate for 'b' by substituting for the value of 'a' in equation 4

b = a - 20 = 109 - 20 = 89.

and to test if the values of a, b and c are correct:

a + b + c = 231

109 + 89 + 33 = 231

4 0
3 years ago
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