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d1i1m1o1n [39]
4 years ago
6

What is the answer to this 177-63

Mathematics
1 answer:
Annette [7]4 years ago
7 0
The answer is 114 to your question
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PLS HELP Find the m∠RLU.<br> A. 10º<br> B. 6º<br> C. 25º<br> D. 50º
balandron [24]

Step-by-step explanation:

6n-10=50

6n=60

n=10

rlu=6n-10

=6×10-10

=60-10

50

8 0
3 years ago
URGENT HELP PLEASE.
AlladinOne [14]
W + 2w = 60

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A can of peas weighs 10 oz. Explain how you would make a graph to model the total weight of peas in terms of the number of cans
Vika [28.1K]
Ur x axis will be the number of cans and ur y axis will be the weight in oz

u will label ur x axis( the number of cans) in intervals of 1......from 0 to 4
u will label ur y axis (the weight) in intervals of 10....from 0 to 40

ur equation would be : y = 10x
points on this line are : (0,0), (1,10), (2,20), (3,30), (4,40)
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3 years ago
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Solve the following system of equations -8x -8y=24 2x + y=-7
leonid [27]

Solve the equation using substitution?

4 0
3 years ago
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Find the annual rate of interest. Principal = 4600 rupees, Period = 5 years, Total amount = 6440 rupees, Annual rate of interest
Len [333]

The annual rate of interest per year is 8%

<u>Solution:</u>

Given:- Principal (p) = 4600 rupees, Time –Period (t) = 5 years, Total amount(A) = 6440 rupees

First we will calculate the Interest and then using formula of simple interest we will calculate the rate of interest

Interest = Amount – Principal

Interest = 6440  – 4600 = 1840

Now using the formula of simple Interest and on putting values we get,

\text {Simple Interest }=\frac{P \times R \times T}{100}

Where "P" is the principal and "R" is the rate of interest per annum and "T" is the time period

1840=\frac{4600 \times R \times 5}{100}

\begin{aligned} \mathrm{R} &=\frac{1840 \times 100}{4600 \times 5} \\\\ \mathrm{R} &=\frac{1840}{230} \\\\ \mathrm{R} &=8 \% \end{aligned}

Hence, the required rate of interest per year is 8%

6 0
3 years ago
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