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Anon25 [30]
3 years ago
14

5-13) -10-15 what’s the equation for the line that passes through points (5,-3) and (-10,15)

Mathematics
1 answer:
umka2103 [35]3 years ago
7 0

Answer: y=-6/5x+3

Step-by-step explanation:

To find the equation of the line that passes through the points, you want to first find the slope by using the formula m=\frac{y_2-y_1}{x_2-x_1}, then solve for the y-intercept.

m=\frac{15-(-3)}{-10-5} =\frac{18}{-15} =-\frac{6}{5}

Now that we have the slope, we can plug in a given point and solve for the y-intercept.

15=-6/5(-10)+b        [multiply]

15=12+b                   [subtract both sides by 12]

b=3

Now that we have the y-intercept, we know the equation is y=-6/5x+3.

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N=11/5

Step-by-step explanation:

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<u>G</u><u>iven </u><u>:</u><u>-</u>

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<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>

  • The value of theta .

<u>Solution</u><u> </u><u>:</u><u>-</u>

At angle theta , 9cm side will be considered as the perpendicular and 10cm side will be hypotenuse . So , as we know that ;

\longrightarrow sin\theta =\dfrac{perpendicular}{hypotenuse}

Substituting the respective values,

\longrightarrow sin\theta =\dfrac{9cm}{10cm}

Simplify,

\longrightarrow sin\theta= 0.9

Take arcsin both sides ,

\longrightarrow \theta =sin^{-1}(0.9)

\\\longrightarrow\underline{\underline{\theta = 64.15^o }}

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6 0
2 years ago
If the distance traveled is (4s3 - 6s² + 4s + 14) miles and the rate is (s + 1) mph, write an expression, in hours, for the time
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The equation for time is URM:

time =  \frac{distance}{speed \: (rate)}

t(s) =  \frac{4s {}^{3}  - 6s {}^{2}  + 4s + 14}{s + 1}  =  \frac{2(s + 1)(2s {}^{2} - 5s + 7) }{s + 1}  \\  t(s) = 2(2s {}^{2}  - 5s + 7) \\ t(s) = 4s {}^{2}  - 10s + 14

3 0
1 year ago
What is the value of 3x^2 + 5x + 25 when x = 3
pshichka [43]

Answer:

Step-by-step explanation:

3x^2 + 5x + 25 when x = 3

= 3(3)^2 + 5(3) + 25

= 3(9) + 15 + 25

= 27 +15 + 25

= 67

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3 years ago
Please help me solve e
irinina [24]

Answer:

A. Positive Slope

Step-by-step explanation:

Positive Slopes have a line going up and to the right

Negative Slopes have a line going to the down and left

0 Slope is a straight horizontal line

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Hope this helps you along your way! :)

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