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3 years ago

I'LL GIVE BRAINLIEST....PLZ HELP ME!!

Mathematics

2 answers:

julsineya [31]3 years ago

5 0

Answer:

(1,7)

Step-by-step explanation:

Since we know that y=5x+2, we replace y in the bottom equation with 5x+2.

3x=-(5x+2)+10

Then distribute the - into both of the parentheses (multiply both terms by -1):

3x=-5x-2+10

Then add like terms:

3x=-5x+8

Add 5x to both sides:

8x=8

Divide both sides by 8:

x=1

Plug the known x value into y=5x+2:

y=5(1)+2

Multiply:

y=5+2

Add:

y=7

Put values into (x,y) form:

(1,7)

Ad libitum [116K]3 years ago

3 0

Answer:

Step-by-step explanation:

y = 5x + 2.....so we will sub in 5x + 2 for y, back into the other equation

3x = -y + 10

3x = - (5x + 2) + 10

3x = -5x - 2 + 10

3x + 5x = 8

8x = 8

x = 8/8

x = 1

y = 5x + 2

y = 5(1) + 2

y = 5 + 2

y = 7

check it... (1,7)

3x = -y + 10 y = 5x + 2

3(1) = -7 + 10 7 = 5(1) + 2

3 = 3 (correct) 7 = 7 (correct)

so ur solution is : x = 1 and y = 7...or (1,7) <=====

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Answer:

$(a)\ \sec^2(\theta) = 82$

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Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

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$\tan(\theta) = \frac{Opposite}{Adjacent}$

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$\frac{Opposite}{Adjacent} = \frac{9}{1}$

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$Hypotenuse =\sqrt{82}$

Solving (a):

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This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

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Where:

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So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

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