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horsena [70]
3 years ago
13

Please help on this question. No links

Mathematics
1 answer:
Dmitrij [34]3 years ago
6 0

Answer:

137

Step-by-step explanation:

m∠B and m∠C are adjacent angles in a parallelogram.

Sum of adjacent angles in a parallelogram is 180.

m∠B + m∠C = 180

43 + m∠C = 180

m∠C = 180 - 43

m∠C = 137

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20 weeks in 1 week

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Step-by-step explanation:

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3 years ago
A principal of $2000 is placed in a savings account at 3% per annum compounded annually. How much is in the account after one ye
natita [175]

Answer:

Step-by-step explanation:

After one year

A=p(1+r/n)^nt

=2000(1+0.03/12)^12*1

=2000(1+0.0025)^12

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=$2060.8

After two-years

A=p(1+r/n)^nt

=2060.8(1+0.03/12)^12*2

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After three years

A=p(1+r/n)^nt

=2188.157(1+0.03/12)^12*3

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8 0
4 years ago
I will mark Brainliest
tamaranim1 [39]

Answer:

Step-by-step explanation:

75^{30}\\\\75=3\cdot5\cdot5=3\cdot5^2\\\\75^{30}=(3\cdot5^2)^{30}=3^{30}\cdot(5^2)^{30}=3^{30}\cdot5^{60}

45^{45}\cdot15^{15}\\\\45=3\cdot3\cdot5=3^2\cdot5\\15=3\cdot5\\\\45^{45}=(3^2\cdot5)^{45}=(3^2)^{45}\cdot5^{45}=3^{90}\cdot5^{45}\\15^{15}=(3\cdot5)^{15}=3^{15}\cdot5^{15}\\\\45^{45}\cdot15^{15}=3^{90}\cdot5^{45}\cdot3^{15}\cdot5^{15}=3^{90+15}\cdot5^{45+15}=3^{105}\cdot5^{60}\\\\=3^{75+30}\cdot5^{60}=3^{75}\cdot\underbrace{3^{30}\cdot5^{60}}_{75^{30}}=3^{75}\cdot75^{30}

\text{therefore it is divisible by}\ 75^{30}

Used:\\\\(ab)^n=a^nb^m\\\\(a^n)^m=a^{nm}\\\\a^n\cdot a^m=a^{n+m}

3 0
3 years ago
Given that the point (9,7) is on the graph of y=f(x), there is one point that must be on the graph of 2y=(f(2x)/2) + 2. What is
natulia [17]

Answer:

Step-by-step explanation:

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4 0
3 years ago
How many 3 number combinations are possible if you can't repeat the numbers? The numbers range from 0-39
TEA [102]

Answer: 49600

Step-by-step explanation: There are 40 ways to choose the first number

and there are 40 ways to choose the second number.

Now we take some examples to determine how many choices there are for

the third number

If the second number is 23, then the 3rd number cannot be any of these

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Let's look at a second example which includes some on both sides of 0.

If second number is 2, we couldn't choose any of the 9 numbers 38,39,0,1,2,3,4,5,6.

So thats 40-9 or 31 choices for the 3rd number.

Answer 40*40*31 = 49600 ways.

4 0
3 years ago
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