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babunello [35]
2 years ago
10

Ryan spent $3.25 on lunch every day, Monday through Friday. If he had $20 at the start of the week, how much money did he have l

eft after Friday
Mathematics
2 answers:
sesenic [268]2 years ago
8 0

Answer:

He had $3.75 dollars left.

Step-by-step explanation:

He was spending launch money for five days so:

3.25*5 is the total amount of money he spent that week.

The amount of Mooney he had minus the amount of money he spent is the amount of money left.

20-3.25*5= 3.75

shusha [124]2 years ago
6 0

Monday through Friday is 5 days.

Multiply the cost of lunch by number of days:

3.25 x 5 = $16.25

Subtract the total he spent on lunch from what he started with for money:

20 - 16.25 = 3.75

He had $3.75 left.

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Answer:

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Step-by-step explanation:

1000*0+321*15678*9(-4563) = -206675344746

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Diego can type 140 in 4 minutes
oksano4ka [1.4K]

Answer:

11 min and 525 words

Step-by-step explanation:

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PLEASE HELP 20 POINTS<br> solve the system of equations y=-5x-7 and -4x-3y=-1 by substitution.
Ainat [17]

Answer:

Solution: x = -2; y = 3 or (-2, 3)

Step-by-step explanation:

<u>Equation 1:</u>  y = -5x - 7  

<u>Equation 2:</u> -4x - 3y = -1

Substitute the value of y in Equation 1 into the Equation 2:

-4x - 3(-5x - 7)  = -1

-4x +15x + 21 = -1

Combine like terms:

11x + 21 =  - 1

Subtract 21 from both sides:

11x + 21 - 21 =  - 1 - 21

11x = -22

Divide both sides by 11 to solve for x:

11x/11 = -22/11

x = -2

Now that we have the value for x, substitute x = 2 into Equation 2 to solve for y:

-4x - 3y = -1

-4(-2) - 3y = -1

8 - 3y = -1

Subtract 8 from both sides:

8 - 8 - 3y = -1 - 8

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Divide both sides by -3 to solve for y:

-3y/-3 = -9/-3

y = 3

Therefore, the solution to the given systems of linear equations is:

x = -2; y = 3 or (-2, 3)

Please mark my answers as the Brainliest if you find this helpful :)

7 0
2 years ago
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Answer:

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Step-by-step explanation:

cos = b/h

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zavuch27 [327]

we know that

the volume of a solid oblique pyramid is equal to

V=\frac{1}{3}*B*h

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B is the area of the base

h is the height of the pyramid

in this problem we have that

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where

<u>b=x\ cm</u>

so

B=x^{2}\ cm^{2}

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V=\frac{1}{3}*[x^{3} +2x^{2}]\ cm^{3}

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