We can use the points (4, 5) and (8, 10) to solve.
Slope formula: y2-y1/x2-x1
= 10-5/8-4
= 5/4
= 1.25
The answer is [ D. The slope of the graph is 1.25 ]
Best of Luck!
Answer:
C and D are correct in my calculations, I'm confused.. ;w;
Step-by-step explanation:
4x + 9 + 79 - x + y = 180
4x - x + y + 79 + 9 = 180
3x + y + 88 = 180
3x + y = 180 - 88
3x + y = 92
y = 92 - 3x
A.) x = 64, y = 50
y = 92 - 3x
(50) = 92 - 3(64)
50 = 92 - 192
50 ≠ - 100
B.) x = 14, y = 152
y = 92 - 3x
(152) = 92 - 3(14)
152 = 92 - 42
152 ≠ 50
C.) x = 23.3, y = 22.1
y = 92 - 3x
(22.1) = 92 - 3(23.3)
22.1 = 92 - 69.9
22.1 = 22.1
D.) x = 14, y = 50
y = 92 - 3x
(50) = 92 - 3(14)
50 = 92 - 42
50 = 50
Answer:
Step-by-step explanation:
The graph crosses the y-axis at (0,-2).
The slope of the graph is 3.
y = 3x - 2
A semicircle is a half of a circle.
Find the circumference of a full circle then divide in half.
Circumference = pi x r x 2
Circumference = 3.14 x 9 x 2 = 56.52
Semicircle = 56.52/2 = 28.26
The original equation:
Integral of csc(5x)
Use a u sub:
u = 5x
du = 5dx
Simplify the du:
![\frac{1}{5} du = dx](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B5%7D%20du%20%3D%20dx)
Apply to the equation:
Integral csc(u)
![\frac{1}{5}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B5%7D%20)
du
Simplify:
Integral
![\frac{1}{5}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B5%7D%20)
![\frac{1}{sin(u)}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bsin%28u%29%7D%20)
du
Factor out the constant:
![\frac{1}{5}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B5%7D%20)
Integral
![\frac{1}{sin(u)}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bsin%28u%29%7D%20)
du
Use a second Substitution:
v = tan(
![\frac{u}{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bu%7D%7B2%7D%20)
)
du =
![\frac{2}{1+v^{2} }](https://tex.z-dn.net/?f=%20%5Cfrac%7B2%7D%7B1%2Bv%5E%7B2%7D%20%7D%20)
dv
Applying to the equation:
![\frac{1}{5} Integral \frac{1}{ \frac{2v}{1+ v^{2} } } * \frac{2}{1+v^{2} } dv](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B5%7D%20Integral%20%20%5Cfrac%7B1%7D%7B%20%5Cfrac%7B2v%7D%7B1%2B%20v%5E%7B2%7D%20%7D%20%7D%20%2A%20%5Cfrac%7B2%7D%7B1%2Bv%5E%7B2%7D%20%7D%20dv)
Simplify:
![\frac{1}{5} Integral \frac{1}{v} dv](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B5%7D%20Integral%20%20%5Cfrac%7B1%7D%7Bv%7D%20dv)
Integrate:
![\frac{1}{5} ln(|v|)](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B5%7D%20ln%28%7Cv%7C%29)
Insert back in your v and u:
v = tan(
![\frac{u}{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bu%7D%7B2%7D%20)
)
u = 5x
This gives us the final equation (don't forget your constant):
![\frac{1}{5} ln(|tan( \frac{5x}{2} )|) + C](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B5%7D%20ln%28%7Ctan%28%20%5Cfrac%7B5x%7D%7B2%7D%20%29%7C%29%20%2B%20C)
(Thank you for making me write it out, I made a mistake on the original answer.)