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Anastaziya [24]
3 years ago
10

When x=12 y=8. find y when x=42

Mathematics
1 answer:
Ahat [919]3 years ago
4 0

Answer:

\frac{12}{8}  =  \frac{42}{y}  \\ y =  \frac{42 \times 8}{12}  \\ \boxed{ y = 28}

<h3>28 is the right answer.</h3>
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What is the number of acute obtuse and right angle of a rectangle
matrenka [14]
The number of a acute angle in a rectangle is 89 through 1 the obtuse angle in a rectangle is 91 through inf and a right angle of a rectangle is 90 degrees 
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3 years ago
Show work and explain with formulas.
Juli2301 [7.4K]

20 Answer: a₁ = 4

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

a_n=324,\ r=3,\ n=5\\\\a_n=a_1 \cdot r^{n-1}\\\\324=a_1\cdot 3^{5-1}\\\\\dfrac{324}{3^4}=a_1\\\\\dfrac{324}{81}=a_1\\\\\large\boxed{4}=a_1

21 Answer: n = 13

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

a_n=\dfrac{1}{64},\ a_1=64,\ r=\dfrac{1}{2}\\\\a_n=a_1 \cdot r^{n-1}\\\\\dfrac{1}{64}=64\cdot \bigg(\dfrac{1}{2}\bigg)^{n-1}\\\\\dfrac{1}{64\cdot 64}=\bigg(\dfrac{1}{2}\bigg)^{n-1}\\\\\dfrac{1}{2^6\cdot 2^6}=\dfrac{1}{2^{n-1}}\\\\6+6=n-1\\\\\large\boxed{13}=n

22 Answer: n = 5

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

a_n=48,\ a_1=1875\ r=\dfrac{2}{5}\\\\a_n=a_1 \cdot r^{n-1}\\\\48=1875\cdot \bigg(\dfrac{2}{5}\bigg)^{n-1}\\\\\dfrac{48}{1875}=\bigg(\dfrac{2}{5}\bigg)^{n-1}\\\\\dfrac{16}{625}=\bigg(\dfrac{2}{5}\bigg)^{n-1}}\\\\\bigg(\dfrac{2}{5}\bigg)^4=\bigg(\dfrac{2}{5}\bigg)^{n-1}}\\\\4=n-1\\\\\large\boxed{5}=n

6 0
3 years ago
Read 2 more answers
I’m doing homework corrections can someone help me with this one
salantis [7]

between 50 and 100 it has 5 lines. so, 100-50 = 50 and divide to 5. because there are 5 parts. 50/5 = 10.1 line = 10. 50 + 10 + 10 = 70.ans: 70

4 0
3 years ago
Find the slope of the line. Write in simplest form<br> Show your work
Marina86 [1]

Answer:

The slope is -2.

Step-by-step explanation:

Following the line starting at the x intercept, move down 2 over right 1.

6 0
3 years ago
Solve for x: 3/3x + 1/x+4 = 10/7x
Mekhanik [1.2K]
Given,

3/3x + 1/(x + 4) = 10/7x

1/x + 1/(x+4) = 10/7x

Because the first term on LHS has 'x' in the denominator and the second term in the LHS has '(x + 4)' in the denominator. So to get a common denominator, multiply and divide the first term with '(x + 4)' and the second term with 'x' as shown below

{(1/x)(x + 4)/(x + 4)} + {(1/(x + 4))(x/x)} = 10/7x

{(1(x + 4))/(x(x + 4))} + {(1x)/(x(x + 4))} = 10/7x

Now the common denominator for both terms is (x(x + 4)); so combining the numerators, we get,

{1(x + 4) + 1x} / {x(x + 4)} = 10/7x

(x + 4 + 1x) / (x(x + 4)) = 10/7x

(2x + 4) / (x(x + 4)) = 10/7x

In order to have the same denominator for both LHS and RHS, multiply and divide the LHS by '7' and the RHS by '(x + 4)'

{(2x+4) / (x(x + 4))} (7 / 7) = (10 / 7x) {(x + 4) / (x + 4)}

(14x + 28) / (7x(x + 4)) = (10x + 40) / (7x(x + 4))

Now both LHS and RHS have the same denominator. These can be cancelled. 

∴14x + 28 = 10x + 40
14x - 10x = 40 - 28
4x = 12
x = 12/4

∴x = 3



8 0
3 years ago
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