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PIT_PIT [208]
3 years ago
14

Wei has $150.00 to make a garland using 60-cent balloons. He wants to purchase 100 blue balloons and some number of white balloo

ns. He learns that the white balloons are on sale for half price. He writes and solves an equation to find the number of white balloons he can purchase.
Mathematics
2 answers:
Montano1993 [528]3 years ago
7 0

Answer:

<h2>It's c</h2>

Step-by-step explanation:

blsea [12.9K]3 years ago
3 0

Answer:

He can purchase 300 white baloons.

Step-by-step explanation:

150$ is 15000 cents

15000 = 60b + 30w

b being number of blue baloons,w number of white baloons and 60 and 30 prices in cents

15000 = 60*100 + 30w

15000 - 60*100 = 30w

15000- 6000 = 30w

30w = 9000

w =300

number of white baloons is 300

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The answer is B. (2x + 5)(x + 1)You could answer this by expanding each answer until you found one that matched 2x^2 + 7x + 5, but I will only show how the answer expands:
2x × x = 2x^2
5 × x = 5x
2x × 1 = 2x
5 × 1 = 5

So in total those brackets expand to 2x^2 + 7x + 5. I hope this helps!
6 0
3 years ago
Plz help me with this question
Alex17521 [72]
Your answer for this question will be the second one on the list
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3 years ago
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Family Video charges $10 for
Irina18 [472]

Answer:

4 movies, $20

Step-by-step explanation:

Given data

Family Video

charges $10 for a monthly membership and

$2.50 per movie

let the number of movies be x

and the charge for x movies be y

y= 10+2.5x----------1

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let the number of movies be x

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y= 12+2x----------2

equate 1 and 2

10+2.5x= 12+2x

collect like terms

12-10= 2.5x-2x

2= 0.5x

x= 2/0.5

x= 4 movies

put x= 4 in equation 1 or 2

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4 0
3 years ago
50 points decreased by 26%
labwork [276]
26% of 50
26/100 * 50
13

50 - 13 = 37


7 0
3 years ago
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What is the 23rd term of the arithmetic sequence where a1 = 8 and a9 = 48?
joja [24]

Answer:

23rd term of the arithmetic sequence is 118.

Step-by-step explanation:

In this question we have been given first term a1 = 8 and 9th term a9 = 48

we have to find the 23rd term of this arithmetic sequence.

Since in an arithmetic sequence

T_{n}=a+(n-1)d

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d = common difference

since 9th term a9 = 48

48 = 8 + (9-1)d

8d = 48 - 8 = 40

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Now T_{23}= a + (n-1)d

= 8 + (23 -1)5 = 8 + 22×5 = 8 + 110 = 118

Therefore 23rd term of the sequence is 118.

4 0
3 years ago
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