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Sever21 [200]
3 years ago
5

Solve the system of equations by substitution. 7+2y=8 3x-2y=0

Mathematics
1 answer:
Volgvan3 years ago
8 0
7+2y=8
3x-2y=0

7+2y=8
y=8/3

3x-2(8/3)=0
3x-16/3=0
+16/3   +16/3
3x=16/3
/3     /3
x=16/9
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Step-by-step explanation:

When the base is even it is positive when the base is odd it is negative

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What is the total number of different 10-letter arrangements that can be formed using the letters in the word FORGETTING?
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1814400

Step-by-step explanation:

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2 years ago
The area of a square is given by x2, where x is the length of one side. Mary's original garden was in the shape of a square. She
Vinil7 [7]

Answer:

128\text{ ft}^{2}

Step-by-step explanation:

We have been given that the area of a square is given by x^2, where x is the length of one side.

Mary's original garden was in the shape of a square. She has decided to double the area of her garden. So the new area of Mary's garden will be 2 times the area of original garden.

We can represent this information in an equation as:

\text{Area of Mary's new garden}=2x^{2}

Therefore, the expression 2x^2 will represent the area of Mary's new garden.

To evaluate the area of new garden, if the side length of Mary's original garden was 8 feet, we will substitute x equals 8 in our expression.

\text{Area of Mary's new garden}=2(8\text{ ft})^{2}

\text{Area of Mary's new garden}=2*64\text{ ft}^{2}

\text{Area of Mary's new garden}=128\text{ ft}^{2}

Therefore, the area of Mary's new garden will be 128 square feet.

4 0
3 years ago
The graph from the US Department of Agriculture shows the growth of farmers markets in the United States. What was the first yea
Helen [10]
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3 0
3 years ago
If A+B+C=<img src="https://tex.z-dn.net/?f=%5Cpi" id="TexFormula1" title="\pi" alt="\pi" align="absmiddle" class="latex-formula"
seraphim [82]

Answer:

a + b + c = \pi \\  =  > c=  \pi - a - b \\  <  =  >  \tan(c)  =  \tan(\pi - a - b)  =  -\tan(a + b)

Step-by-step explanation:

we have:

\tan(a)  +  \tan(b)  +  \tan(c)  \\  =  \tan(a)  +  \tan(b)  -  \tan(a + b)  \\  =  \tan( a)  +  \tan(b)  -  \frac{ \tan(a) +  \tan(b)  }{1 -  \tan(a)  \tan(b) }  \\  =  \frac{ ( \tan(a) +  \tan(b)  ) \tan(a) \tan(b)  }{ \tan(a) \tan(b)  - 1 } (1)

we also have:

\tan(a)  \tan(b)  \tan(c)  \\  =  -  \tan(a)  \tan(b)  \tan(a + b)  \\  =  \frac{ -(\tan( a  )   + \tan(b) ) \tan(a)  \tan(b) }{1 -  \tan(a)  \tan(b) }  \\  =  \frac{( \tan(a)  +  \tan(b)) \tan(a)   \tan(b) }{ \tan(a) \tan(b)  - 1 } (2)

from (1)(2) => proven

5 0
3 years ago
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