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docker41 [41]
3 years ago
14

(with solution)

Mathematics
1 answer:
Lunna [17]3 years ago
8 0

Answer:

see explanation

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + b ( m is the slope and c the y- intercept )

(1)

Here m = 6 and b = - 2, then

y = 6x - 2

(2)

The equation of a line in standard form is

Ax + By = C ( A is a positive integer and B, C are integers )

Here m = - 2 and b = 5, then

y = - 2x + 5 ← equation in slope- intercept form

Add 2x to both sides

2x + y = 5 ← equation in standard form

(3)

Calculate the slope m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (6, - 13) and (x₂, y₂ ) = (- 4, - 3)

m = \frac{-3+13}{-4-6} = \frac{10}{-10} = - 1 , then

y = - x + b ← is the partial equation

To find b substitute either of the 2 points into the partial equation

Using (- 4, - 3 ) , then

- 3 = 4 + b ⇒ b = - 3 - 4 = - 7

y = - x - 7 ← equation in slope- intercept form

Add x to both sides

x + y = - 7 ← equation in standard form

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Answer:

42 + 6w

Step-by-step explanation:

Basically,

6(8+w) is 6 x (8+w)

OR 6(8) + 6(w)

Hence 42 + 6w

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Barrel A and Barrel B are the same size. Barrel A is filled with a solution that is 12% ethanol. The portion of ethanol in the s
mario62 [17]

Answer:

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Step-by-step explanation:

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Katelyn had 2 dogs and 3 cats. Which of the following shows the ratio written correctly for the number of dogs to cats?
Rudik [331]

Answer:

2:3

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If tan theta = 10/13 and cos theta > 0 then sin2theta is what ?
svetoff [14.1K]

sin2\alpha  = \frac{260}{269}

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

We have , Tan\alpha  = \frac{Perpendicular}{Base} = \frac{10}{13},

We know that sin\alpha  = \frac{Perpendicular}{Hypotenuse} = \frac{Perpendicular}{\sqrt[2]{(Perpendicualr)^{2} + (Base)^{2})} }

Substituting values of P & B , sin\alpha  = \frac{10}{\sqrt{10^{2} + 13^{2}} }  = \frac{10}{\sqrt{269} }

Now , sin2\alpha  = 2sin\alpha cos\alpha  = 2sin\alpha \sqrt{1 - (sin\alpha)^{2} }

⇒sin2\alpha  = \frac{10}{\sqrt{269} } ×\sqrt{1 - (\frac{10}{\sqrt{269} })^{2} }×2

⇒ sin2\alpha  = \frac{20}{\sqrt{269} }( \sqrt{\frac{269 - 100}{269} }  )

⇒sin2\alpha  = \frac{20}{\sqrt{269} }( \sqrt{\frac{169 }{269} }  )

⇒sin2\alpha  = \frac{260}{\sqrt{269} }( \sqrt{\frac{1}{269} }  )

⇒sin2\alpha  = \frac{260}{269}

6 0
3 years ago
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