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Artyom0805 [142]
2 years ago
14

Every evening Jenna empties her pockets and puts her change in a jar. At the end of the week she counts her money. One week she

had 40 coins, all of them dimes and quarters. When she added them up, she had a total of $7.75. Let d= number of dimes and q= number of quarters
Mathematics
1 answer:
Sergeeva-Olga [200]2 years ago
6 0
A) .10 d + .25 q = 7.75
B) d + q = 40

Multiplying B) by -.10
B)  -.10d -.10q = -4.0
Then adding this to A)
A) .10 d + .25 q = 7.75
.15q = 3.75

Quarters = 25
Therefore, dimes = 15
***************************************************
Double-Check
A) .10 d + .25 q = 7.75
A) .10 * 15 d + .25 * 25 = 7.75
A) 1.50 + 6.25 = 7.75

Correct!!


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4 0
2 years ago
Imagine that these are test scores on a Statistics examinations (test):
yanalaym [24]

Answer:

a

Step-by-step explanation:

7 0
2 years ago
Share 800 in the ratio 9:7.
bulgar [2K]

<u>Answer:</u>

  • The numerator '9's share is 450 and the denominator '7's share is 350.

<u>Step-by-step explanation:</u>

<u>Step-1: Add the ratios:</u>

  • 9 + 7
  • 16

<u>Step-2: Divide 800 by the sum of the ratios:</u>

  • 800 ÷ 16
  • => 50

<u>Step-3: Multiply 50 with each ratio:</u>

  • 9 x 50 = 450
  • 7 x 50 = 350

Hence, <u>the numerator '9's share is 450 and the denominator '7's share is 350.</u>

Hoped this helped.

BrainiacUser1357

<u />

8 0
2 years ago
Let a = x2 + 4. Use a to find the solutions for the following equation:
Zepler [3.9K]

The solutions for x are -2, 0, 2

<em><u>Solution:</u></em>

Given that,

\text { Let } a = x^2 + 4

Given equation is:

(x^2 + 4)^2 + 32 = 12x^2 + 48

(x^2 + 4)^2 + 32 = 12(x^2 + 4)

\text { Subtsitute } a = x^2 + 4 \text{ in above equation }

a^2 + 32 = 12a\\\\a^2 -12a + 32 = 0

\text{ Let us factorize the above equation }\\\\\text{ Splitting the middle term -12a as -4a - 8a we get, }

a^2 -4a - 8a + 32 = 0

Taking "a" as common term from first two terms and taking "-8" as common from last two terms

a(a-4)-8(a - 4) = 0

\text{Taking (a - 4) as common term, }\\\\(a - 4)(a - 8) = 0

\text{Equating to zero we get, }\\\\(a - 4) = 0 \text{ or } (a - 8) = 0\\\\a = 4 \text{ or } a = 8

\text{Now substitute the value of a = 8 and a = 4 in: }\\\\a = x^2 + 4

\text{ For a = 8: }\\\\a = x^2 + 4\\\\8 = x^2 + 4\\\\x^2 = 4\\\\x = \pm2\\\\x = +2 \text{ or } -2

\text{For a = 4: }\\\\a = x^2 + 4\\\\4 = x^2 + 4\\\\x = 0

\text{Thus solutions of } x \text{ are: }\\\\x = 0 \text{ or } x = 2 \text{ or } x = -2

8 0
3 years ago
Helppp can u finish the problem plz show your work!!!!
sveticcg [70]
_
2.6
__
3/8.0
-6
-------
20
- 18
--------
2

_
answer: 2.6
4 0
3 years ago
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