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Bezzdna [24]
3 years ago
6

Factor completely, then place the factors in the proper location on the grid.

Mathematics
1 answer:
Lyrx [107]3 years ago
6 0
4(x+5y)(x-3y)
I don't know what you mean by placing the factors on the grid
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You keep the same exponent when adding or subtracting
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Katie buys a large pizza for $14.99 and bottles of soda for $1.50 each. If the total cost was $23.99, how many bottles of soda (
Nady [450]

The number of soda bottles Katie bought is 6

The given parameters are:

  • Large Pizza = $14.99
  • Soda = $1.50 per bottle
  • Total cost = $23.99

Assume the number of soda bottle is x.

So, we have:

Total = Pizza + Soda \times x

This gives

23.99 = 14.99 + 1.50\times x

Subtract 14.99 from both sides of the equations

9 = 1.50\times x

Divide both sides of the equations by 1.5

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Rewrite the equation as

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Hence, the number of bottles is 6

Read more about linear equations at:

brainly.com/question/14323743

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2 years ago
Find the slope of line that passes through (5,6) and (7,9)
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Step-by-step explanation:

m=\frac{9-6}{7-5}

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5 0
3 years ago
Disks of polycarbonate plastic from a supplier are analyzed for scratch and shock resistance. The results from 100 disks are sum
Naddika [18.5K]

Answer:

P(A) = 86/100

P(B) = 79/100

P(A') = 14/100

P(A U B) = 95/100

P(A' U B) = 84/100

Step-by-step explanation:

i. P(A) = number of disks with high shock resistance/total number of disks

= (70+16)/100

= 86/100

ii. P(B) = number of disks with high scratch resistance/total number of disks

= (70+9)/100

= 79/100

iii. P(A') = number of disks without high shock resistance/total number of disks

= (100-86)/100

= 14/100

iv. P(AUB) = number of disks with high shock resistance or high scratch resistance/total number of disks

= (70+9+16)/100

= 95/100

v. P(A'UB) = number of disks without high shock resistance or with high scratch resistance/total number of disks

= (5+9+70)/100

= 84/100

3 0
3 years ago
F(x) = 4*<br> How to evaluate
photoshop1234 [79]

Answer:fx=4

Step-by-step explanation:

fx=4

7 0
3 years ago
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