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sashaice [31]
3 years ago
15

zasha spent $6 on packages of gum.How many more packages of gum that cost $1.20 each can she buy if she has a $20 bill

Mathematics
1 answer:
antiseptic1488 [7]3 years ago
4 0
You have to make an inequality:
1.2x + 6 ≤ 20
       - 6     - 6     isolate x

1.2x ≤ 14
÷1.2     ÷1.2   isolate x

x ≤  11.666

Since you can't buy 2/3 of a pack, she can only buy 11 pieces of gum.

Hope this helps!
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The factorization of 8x3 -125 is (2x-5)(jx2 +kx+25)
lbvjy [14]

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3 years ago
Weekly wages at a certain factory
shtirl [24]
Where is the question
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3 years ago
Write an equation in slope-intercept form for the line that satisfies the following condition. slope 7, and passes through (5, 3
allsm [11]
Y = mx + b
slope(m) = 7
(5,30)...x = 5 and y = 30
now we sub and find b, the y int
30 = 7(5) + b
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7 0
3 years ago
Use a Venn diagram to answer the question. A survey of 180 families showed that 67 had a​ dog; 52 had a​ cat; 22 had a dog and a
Mice21 [21]

Answer:

There are 13 families had a parakeet only

Step-by-step explanation:

* Lets explain the problem

- There are 180 families

- 67 families had a dog

- 52 families had a cat

- 22 families had a dog and a cat

- 70 had neither a cat nor a​ dog, and in addition did not have a​

 parakeet

- 4 had a​ cat, a​ dog, and a parakeet (4 is a part of 22 and 22 is a part

 of 67 and 520

* We will explain the Venn-diagram

- A rectangle represent the total of the families

- Three intersected circles:

 C represented the cat

 D represented the dog

 P represented the parakeet

- The common part of the three circle had 4 families

- The common part between the circle of the cat and the circle of the

 dog only had 22 - 4 = 18 families

- The common part between the circle of the dog and the circle of the

 parakeet only had a families

- The common part between the circle of the cat and the circle of the

 parakeet only had b families

- The non-intersected part of the circle of the dog had 67 - 22 - a =

  45 - a families

  had dogs only

- The non-intersected part of the circle of the cat had 52 - 22 - b =  

  30 - b families

  had cats only

- The non-intersected part of the circle of the parakeet had c families

  had parakeets only

- The part out side the circles and inside the triangle has 70 families

- Look to the attached graph for more under stand

∵ The total of the families is 180

∴ The sum of all steps above is 180

∴ 45 - a + 18 + 4 + 30 - b + b + c + a + 70 = 180 ⇒ simplify

- (-a) will cancel (a) and (-b) will cancel (b)

∴ (45 + 18 + 4 + 30 + 70) + (-a + a) + (-b + b) + c = 180

∴ 167 + c = 180 ⇒ subtract 167 from both sides

∴ c = 180 - 167 = 13 families

* There are 13 families had a parakeet only

5 0
3 years ago
A pile of coins, consisting of quarters and half dollars, it worth $11.75. If there are 2 more quarters than half dollars, how m
lord [1]
Let the number of half dollars be x,

number of quarters = x + 2
amount of half dollars = 0.5x
amount of quarters = 0.25(x + 2)
= 0.25x + 0.5
total amount = 0.5x + 0.25x + 0.5
= 0.75x + 0.5
0.75x + 0.5 = 11.75
0.75x = 11.25
x = 15
number of quarters = x + 2
= 15 + 2
= 17

There are 17 quarters and 15 half dollars.
3 0
3 years ago
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