Answer:
From his starting point of 85 3/4, he descended 17 9/20 meters
Step-by-step explanation:
We want to know how much farther he went down from his starting point.
103 1/5 - 85 3/4
S
tart off by making the denominators of both the numbers the same.
103 1/5 - 85 3/4
103 4/20 - 85 15/20
Make them into improper fractions
2064/20 - 1715/20 = 349/20
Make 349/20 into a mixed fraction
17 9/20
solution:
we know that ,
u.v = ΙuΙ ΙvΙcosθ
here,
θ =60° (since the given triangle is equilateral triangle)
u.v = ΙuΙ ΙvΙcos60°
= 1 x 1 x 1/2
u.v = 1/2
now, u.w = ΙuΙ ΙwΙcosθ
= ΙuΙ x cos(60x2)
u.w = -1/2
Your answer would be option B. 2y² - y - 6 = 0. This is because if you were to substitute x = y² - 1 into the equation 2x - y = 4, you would get 2(y² - 1) - y = 4, which expands into 2y² - 2 - y = 4, and then simplifies to 2y² - y - 6 = 0.
I hope this helps!
9514 1404 393
Answer:
C. √((5-9)² +(1+6)²)
Step-by-step explanation:
The distance formula can be written ...
d = √((x1 -x2)² +(y1 -y2)²)
Filling in the given point coordinates, this becomes ...
d = √((5 -9)² +(1 -(-6))²)
Simplifying the signs, this becomes ...
d = √((5 -9)² +(1 +6)²) . . . . . matches choice C
If f(x) has an inverse on [a, b], then integrating by parts (take u = f(x) and dv = dx), we can show
Let 
. Compute the inverse:
![f\left(f^{-1}(x)\right) = \sqrt{1 + f^{-1}(x)^3} = x \implies f^{-1}(x) = \sqrt[3]{x^2-1}](https://tex.z-dn.net/?f=f%5Cleft%28f%5E%7B-1%7D%28x%29%5Cright%29%20%3D%20%5Csqrt%7B1%20%2B%20f%5E%7B-1%7D%28x%29%5E3%7D%20%3D%20x%20%5Cimplies%20f%5E%7B-1%7D%28x%29%20%3D%20%5Csqrt%5B3%5D%7Bx%5E2-1%7D)
and we immediately notice that ![f^{-1}(x+1)=\sqrt[3]{x^2+2x}](https://tex.z-dn.net/?f=f%5E%7B-1%7D%28x%2B1%29%3D%5Csqrt%5B3%5D%7Bx%5E2%2B2x%7D)
.
So, we can write the given integral as
Splitting up terms and replacing 
in the first integral, we get
