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PilotLPTM [1.2K]
3 years ago
15

When buying snacks for her classroom, Ms. Davis bought 4 bags of mixed nuts and 8 bags of pretzels. A bag of pretzels costs $1.5

0 less than a bag of mixed nuts. She spent $57 for the snacks. The variable is the price of one bag of mixed nuts.
What is the translation of this scenario?
A cost of pretzels = cost of mixed nuts
B cost of pretzels – cost of mixed nuts = 57
C cost of mixed nuts + cost of pretzels = 57
D 57 + cost of mixed nuts = cost of pretzels
Mathematics
2 answers:
adell [148]3 years ago
7 0

Answer:

c

Step-by-step explanation:

vfiekz [6]3 years ago
3 0

Answer:

C

Step-by-step explanation:

The snacks altogether (Pretzels and Nuts) cost 57 dollars.

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6 0
3 years ago
Read 2 more answers
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
prohojiy [21]

Answer:

2.25

Step-by-step explanation:

(6x^{-2})^2(0.5x)^4\\\\=(6^2(x^{-2})^{2})(0.5^4x^4)\ \ \ \ \ \ \ \ \ \ \ \ \ as\ (ab)^m=a^mb^m\\\\=36\times 0.5^4((x^{-2})^2x^4)\ \ \ \ \ \ \ \ \ \ as\ multiplication\ is\ associative\ a(bc)=(ab)c\\\\=36\times 0.0625(x^{-4}x^4)\ \ \ \ \ \ \ \ \ \ \ as\ (x^m)^n=x^{mn}\\\\=(36\times 0.0625)(x^{-4+4})\ \ \ \ \ \ \ \ \ \ \ as\ x^mx^n=x^{m+n}\\\\=2.25x^0\\\\=2.25\ \ \ \ \ \ \ \ \ \ \ \ as\ x^0=1\\\\(6x^{-2})^2(0.5x)^4=2.25

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2 years ago
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hoa [83]

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8 0
3 years ago
Find the solution of y = 6x + 1 for x = 5.
user100 [1]
Y=6x+1
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y=31
D. (5, 31)
hope this helps!
4 0
3 years ago
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