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Marysya12 [62]
2 years ago
7

Need help with question

Mathematics
1 answer:
svetoff [14.1K]2 years ago
5 0

Answer:

15 14/15

Step-by-step explanation:

8 + 7 = 15

1/3 + 3/5

5/15 + 9/15

14/15

15 + 14/15 = 15 14/15

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tonya earned $5.00 she put half of the money into savings and kept the rest to spend.she bought a toy for $2.00.how much spendin
nalin [4]

Answer:

50 cents

Step-by-step explanation:

half of 5.00$ is 2.50$

6 0
3 years ago
Can you help me with #6???
AnnZ [28]
What are you asking?
6 0
3 years ago
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Building a is 160160 feet shorter than building
Serjik [45]
I assume the heights are 160 ft and 1480 ft.

The two heights are unknown, so we will use variable h to help solve the problem.
The shorter building, building A, has height h.
Since building A is shorter by 160 ft, then building B is taller by 160 ft, so the height of building B is h + 160.

Now we add our two heights to find the total height.

h + h + 160 is the total height.
We can write it as 2h + 160

We are told the total height is 1480 ft, so we let 2h + 160 equal 1480, and we have an equation.

2h + 160 = 1480

Subtract 160 from both sides

2h = 1320

Divide both sides by 2

h = 660

h + 160 = 820

Building A measures 660 ft.
building B measures 820 ft.
7 0
3 years ago
This is big ideas work pls show how u do it so i know
Komok [63]

Answer:

(5,-1)

Step-by-step explanation:

7 0
2 years ago
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(5x^3-8x^2+9x+12)/(x-3)<br><br> please answer with sentences on how to do it step by step. thanks!
Anastaziya [24]

Answer:

5x^2+7x+30+\frac{102}{x-3}

Step-by-step explanation:

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}5x^3-8x^2+9x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{5x^3}{x}=5x^2\\\mathrm{Quotient}=5x^2\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}5x^2:\:5x^3-15x^2\\\mathrm{Subtract\:}5x^3-15x^2\mathrm{\:from\:}5x^3-8x^2+9x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=7x^2+9x+12\\\mathrm{Therefore}\\=5x^2+\frac{7x^2+9x+12}{x-3}

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}7x^2+9x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{7x^2}{x}=7x\\\mathrm{Quotient}=7x\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}7x:\:7x^2-21x\\\mathrm{Subtract\:}7x^2-21x\mathrm{\:from\:}7x^2+9x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=30x+12\\\mathrm{Therefore}\\=5x^2+7x+\frac{30x+12}{x-3}\\

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}30x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{30x}{x}=30\\\mathrm{Quotient}=30\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}30:\:30x-90\\\mathrm{Subtract\:}30x-90\mathrm{\:from\:}30x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=102\\\mathrm{Therefore}\\=5x^2+7x+30+\frac{102}{x-3}

6 0
2 years ago
Read 2 more answers
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