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3 years ago

Write the equation of the line that goes through points (9, 1) and (-1, 6) in slope-intercept form.

Mathematics

1 answer:

stepladder [879]3 years ago

7 0

Answer:

y= -1/2x + 11/2

Step-by-step explanation:

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What is the domain of this function

Talja [164]

Answer:

I think its A

Step-by-step explanation:

8 0

2 years ago

Please help me for my finals!! I'll do anything

oee [108]

(A)

Choice (B) is not the right answer because I

it's only 3rd order. But because f(x) goes to negative infinity when x goes to positive infinity means that the right answer is (A).

6 0

3 years ago

Rose is drawing a portrait for her art class. She is using a sheet of paper what is 6 inches wide and 12 inches long. What is th

rosijanka [135]

Answer:

0.5 ft^2

Step-by-step explanation:

The area in square inches is (6 in )(12 in) = 72 in^2.

Recall that 1 ft^2 = 144 in^2.

Then

72 in^2 1 ft^2

--------------- * --------------- = 0.5 ft^2

1 144 in^2

6 0

3 years ago

The slope-intercept form of the equation of a line that passes through point (-2,-13) is y=5x-3. What is the point-slope form of

marissa [1.9K]

Answer:

y + 13 = 5(x + 2)

Step-by-step explanation:

<h3>Point - slope form of equation: </h3>

$\sf \boxed{\bf y - y_1=m(x-x_1)}$

Slope intercept of the line : y =5x - 3

Comparing with y = mx +b , m =slope = 5

m = 5 and Point (-2 ,-13)

y - [-13] = 5(x - [-2])

y + 13 = 5(x + 2)

4 0

2 years ago

Prove that :

Dimas [21]

Answer:

<h3>Question 1</h3>

sec²θ + cosec²θ =
1/cos²θ + 1/sin²θ =
(sin²θ + cos²θ)/(sin²θcos²θ) =
1 / (sin²θcos²θ) =
[(sin²θ + cos²θ)/sinθcosθ]² =
(sinθ/cosθ + cosθ/sinθ)² =
(tanθ + cotθ)²

<h3>Question 2</h3>

(1 - tan²θ) / (1 + tan²θ) =
(1 - sin²θ/cos²θ) / (1 + sin²θ/cos²θ) =
(cos²θ - sin²θ) / (cos²θ + sin²θ) =
(cosθ + sinθ)(cosθ - sinθ) / 1 =
(cosθ + sinθ)(cosθ - sinθ)

<h3>Question 3</h3>

sinθ/ (1 - cotθ) + cosθ / (1 - tanθ) =
sinθ / (1 - cosθ/sinθ) + cosθ / (1 - sinθ/cosθ) =
sinθ/ [(sinθ - cosθ) / sinθ] + cosθ / [(cosθ - sinθ)/cosθ] =
sin²θ/ (sinθ - cosθ) + cos²θ/(cosθ - sinθ) =
sin²θ/ (sinθ - cosθ) - cos²θ/(sinθ - cosθ) =
(sin²θ - cos²θ) / (sinθ - cosθ) =
(sinθ + cosθ)(sinθ - cosθ) / (sinθ - cosθ) =
sinθ + cosθ

6 0

3 years ago

Read 2 more answers