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irinina [24]
4 years ago
5

The line that passes through the points (9,1) and (5,5) intersects the y-axis at (0,5).

Mathematics
1 answer:
Ilia_Sergeevich [38]4 years ago
6 0
The stament are true I think
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Point A (12, 6) and point B(8,4), find the distance between these two points
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3 years ago
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Write the standard form of a line that passes through (1, 1) and (3, 4)
Sophie [7]

Answer:

3x - 2y = 1

Step-by-step explanation:

Step 1:   Find the slope

slope m = (y2 - y1)(/x2 - x1)

 m = (4 - 1)/(3 - 1) = 3/2

Step 1:  Use the formula y = mx + b to solve

This is slope-intercept form of a line.  We have a slope of 3/2, and a point (x, y), which is (1, 1).  Plug those values into the equation above and solve for b (the only variable we are missing

1 = (3/2)(1) + b

   1 = 3/2 + b  

          1 - 3/2 = b

             2/2 - 3/2 = b

                  -1/2 = b

Step 2: Rewrite the formula using the slope and the b value we just calculated

 y = (3/2)x - 1/2

Step 3:  Standard form of a line is ax + by = c, where a is a positive integer, so we rearrange the equation from step 2 to standard form.  

y = (3/2)x - 1/2

 -(3/2)x + y = -1/2     (subtract (3/2)x from both sides)

 

      -3x + 2y = -1       (multiply by 2 to get rid of the fraction on x)

           3x - 2y = 1     (multiply by -1 so a becomes positive)    

3 0
3 years ago
M=2 (-1, -4) slope intersect form
aalyn [17]
y=mx+b\\\\
-4=2\cdot(-1)+b\\
-4=-2+b\\
b=-2\\\\
\boxed{y=2x-2}
8 0
3 years ago
Cos(theta) = - 3/4 and is in the 3rd quadrant, find the following:
elena-14-01-66 [18.8K]

Answer:

\begin{gathered} sin(\theta)=\frac{-\sqrt{7}}{4} \\ cos(\theta)=-\frac{3}{4} \\ tan(\theta)=\frac{\sqrt{7}}{3} \\ csc(\theta)=-\frac{4}{\sqrt{7}} \\ sec(\theta)=-\frac{4}{3} \\ cot(\theta)=\frac{3}{\sqrt{7}} \end{gathered}

Step-by-step explanation:

If theta is in the third quadrant, draw the diagram to easily identify the other trigonometric relations:

Solve for the missing leg of the triangle, using the Pythagorean theorem:

\begin{gathered} \text{ adjacent}^2+\text{ opposite}^2=\text{ hypotenuse}^2 \\ -3^2+\text{ opposite}^2=4^2 \\ \text{ opposite=}\sqrt{16-9} \\ \text{ opposite=}\sqrt{7} \end{gathered}

Therefore, for the trigonometric relationships:

\begin{gathered} \text{ sin\lparen}\theta)=\frac{opposite}{hypotenuse} \\ \text{ cos\lparen}\theta)=\frac{adjacent}{hypotenuse} \\ tan(\theta)=\frac{opposite}{adjacent} \\ csc(\theta)=\frac{hypotenuse}{opposite} \\ sec(\theta)=\frac{hypotenuse}{adjacent} \\ cot(\theta)=\frac{adjacent}{opposite} \end{gathered}

Now, substitute and solve for the relations:

\begin{gathered} sin(\theta)=\frac{-\sqrt{7}}{4} \\ cos(\theta)=-\frac{3}{4} \\ tan(\theta)=\frac{\sqrt{7}}{3} \\ csc(\theta)=-\frac{4}{\sqrt{7}} \\ sec(\theta)=-\frac{4}{3} \\ cot(\theta)=\frac{3}{\sqrt{7}} \end{gathered}

7 0
1 year ago
Simplify the radical: √80
Nastasia [14]

Answer:

4√5

Step-by-step explanation:

if this isnt the answer u needed lmk!!

7 0
3 years ago
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