I need more info to help you on this
The Tangent Line Problem 1/3How do you find the slope of the tangent line to a function at a point Q when you only have that one point? This Demonstration shows that a secant line can be used to approximate the tangent line. The secant line PQ connects the point of tangency to another point P on the graph of the function. As the distance between the two points decreases, the secant line becomes closer to the tangent line.
I started by labeling the right angle (Angle C) 90º. Next, I wrote down everything in one equation.
2x + 90 + 3x - 20 = 180º (180 degrees in a triangle)
Next, I add 20 on both sides.
2x + 90 + 3x = 200º
I combine like terms (2x and 3x)
5x + 90 = 200º
I subtract 90 from both sides.
5x = 110º
Divide 110 by 5 to get x.
x = 22
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For problem two, I label all the angles I know.
49º + 80º + r = 180º
I add 80 and 49.
129º + r = 180º
I subtract 180 and 129 and get 51º, which is your angle for R.
For angle X, you know that angle R plus angle X equals half of a circle, which is 180º
We know from before that 129º is 180º without R, so X is 129º
I hope this helps! Let me know if I'm wrong!
Answer:
8/11
Step-by-step explanation:
2 2/3 = 8/3
8/3 x 3/11 = 24/33
24/33 = 8/11
Answer:
P= 20.2 and A= 15.675
Step-by-step explanation:
The formula for the area of a triangle is 1/2 base*height. Therefore, the equation for the area of the triangle is 1/2*5.7*5.5. 1/2*5.7=2.85. 2.85*5.5=15.675(area). To find perimeter you're simply adding the numbers. When you add up the numbers of the triangle, you get 20.2.
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