Answer:
y = 5x -12
Step-by-step explanation:
point-slope form:
y - y1 = m(x-x1)
m= slope
m= (y2-y1)/ (x2-x1)
we have (4, 8) and (2,-2)
x1 = 4 y1= 8
x2= 2 y2= -2
m=( -2-8) / (2- 4)
m= -10/ -2
m= 5
so we have:
y - 8 = 5(x-4)
y - 8= 5x -20
y= 5x -20 +8
y = 5x -12
Answer:
Step-by-step explanation:

<h2 /><h2>
<u>Consider</u></h2>

<h2>
<u>W</u><u>e</u><u> </u><u>K</u><u>n</u><u>o</u><u>w</u><u>,</u></h2>




So, on substituting all these values, we get




<h2>Hence,</h2>

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<h2>ADDITIONAL INFORMATION :-</h2>
Sign of Trigonometric ratios in Quadrants
- sin (90°-θ) = cos θ
- cos (90°-θ) = sin θ
- tan (90°-θ) = cot θ
- csc (90°-θ) = sec θ
- sec (90°-θ) = csc θ
- cot (90°-θ) = tan θ
- sin (90°+θ) = cos θ
- cos (90°+θ) = -sin θ
- tan (90°+θ) = -cot θ
- csc (90°+θ) = sec θ
- sec (90°+θ) = -csc θ
- cot (90°+θ) = -tan θ
- sin (180°-θ) = sin θ
- cos (180°-θ) = -cos θ
- tan (180°-θ) = -tan θ
- csc (180°-θ) = csc θ
- sec (180°-θ) = -sec θ
- cot (180°-θ) = -cot θ
- sin (180°+θ) = -sin θ
- cos (180°+θ) = -cos θ
- tan (180°+θ) = tan θ
- csc (180°+θ) = -csc θ
- sec (180°+θ) = -sec θ
- cot (180°+θ) = cot θ
- sin (270°-θ) = -cos θ
- cos (270°-θ) = -sin θ
- tan (270°-θ) = cot θ
- csc (270°-θ) = -sec θ
- sec (270°-θ) = -csc θ
- cot (270°-θ) = tan θ
- sin (270°+θ) = -cos θ
- cos (270°+θ) = sin θ
- tan (270°+θ) = -cot θ
- csc (270°+θ) = -sec θ
- sec (270°+θ) = cos θ
- cot (270°+θ) = -tan θ
Answer:
17-2x=-11 ---> 14
Step-by-step explanation:
plug in the "answers"
17-2(14)=-11
Answer:
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Step-by-step explanation:
From the two right triangles, you can write the following equations using the Pythagorean theorem. Let's call that shared leg in the middle "y"
y^2 + b^2 = a^2
y^2 + c^2 = x^2
y^2 + b^2 = a^2
re-write this to get "y" alone for substitution.
y^2 = a^2 - b^2
substitute (a^2 - b^2) for y^2 in the other equation. y^2 + c^2 = x^2
a^2 - b^2 + c^2 = x^2
Now put in the values given for a,b,c to solve for x
(7.1)^2 - (5.6)^2 + (5.7)^2 = x^2
51.54 = x^2
square root
7.2 = x