The location of the point R will be at ![R(x, y) =(\frac{-32}{5}, \frac{12}{5} )](https://tex.z-dn.net/?f=R%28x%2C%20y%29%20%3D%28%5Cfrac%7B-32%7D%7B5%7D%2C%20%5Cfrac%7B12%7D%7B5%7D%20%20%29)
The midpoint formula is expressed according to the formula;
![R(x, y) = (\frac{ax_1+bx_2}{a+b},\frac{ay_1+by_2}{a+b})](https://tex.z-dn.net/?f=R%28x%2C%20y%29%20%3D%20%28%5Cfrac%7Bax_1%2Bbx_2%7D%7Ba%2Bb%7D%2C%5Cfrac%7Bay_1%2Bby_2%7D%7Ba%2Bb%7D%29)
where;
Given the coordinates Q(-8, 0) and S(12, 0) partitioned in a 4:1 ratio, the coordinate of point R will be expressed as:
![R(x, y) = (\frac{4(-8)+0}{4+1},\frac{1(12)}{4+1})\\R(x, y) =(\frac{-32}{5}, \frac{12}{5} )](https://tex.z-dn.net/?f=R%28x%2C%20y%29%20%3D%20%28%5Cfrac%7B4%28-8%29%2B0%7D%7B4%2B1%7D%2C%5Cfrac%7B1%2812%29%7D%7B4%2B1%7D%29%5C%5CR%28x%2C%20y%29%20%3D%28%5Cfrac%7B-32%7D%7B5%7D%2C%20%5Cfrac%7B12%7D%7B5%7D%20%20%29)
Hence the location of the point R will be at ![R(x, y) =(\frac{-32}{5}, \frac{12}{5} )](https://tex.z-dn.net/?f=R%28x%2C%20y%29%20%3D%28%5Cfrac%7B-32%7D%7B5%7D%2C%20%5Cfrac%7B12%7D%7B5%7D%20%20%29)
Learn more on midpoint here: brainly.com/question/5566419
Answer:
All I know for a fact is that number 10 is C
Step-by-step explanation:
Answer:
1.125
Step-by-step explanation:
729/512=1.423828125
3√1.423828125=1.125
Answer:
∫
= ![\frac{1}{2}(x^6-2)^2+C](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28x%5E6-2%29%5E2%2BC)
Step-by-step explanation:
To find:
∫![6x^5(x^6-2)\,dx](https://tex.z-dn.net/?f=6x%5E5%28x%5E6-2%29%5C%2Cdx)
Solution:
Method of substitution:
Let ![x^6-2=t](https://tex.z-dn.net/?f=x%5E6-2%3Dt)
Differentiate both sides with respect to ![t](https://tex.z-dn.net/?f=t)
![6x^5\,dx=dt](https://tex.z-dn.net/?f=6x%5E5%5C%2Cdx%3Ddt)
[use
]
So,
∫
= ∫
=
where
is a variable.
(Use ∫
)
Put ![t=x^6-2](https://tex.z-dn.net/?f=t%3Dx%5E6-2)
∫
= ![\frac{1}{2}(x^6-2)^2+C_1](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28x%5E6-2%29%5E2%2BC_1)
Use ![(a-b)^2=a^2+b^2-2ab](https://tex.z-dn.net/?f=%28a-b%29%5E2%3Da%5E2%2Bb%5E2-2ab)
So,
∫
= ![\frac{1}{2}(x^6-2)^2+C_1=\frac{1}{2}(x^{12}+4-4x^6)+C_1=\frac{x^{12} }{2}-2x^6+2+C_1=\frac{x^{12} }{2}-2x^6+C](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28x%5E6-2%29%5E2%2BC_1%3D%5Cfrac%7B1%7D%7B2%7D%28x%5E%7B12%7D%2B4-4x%5E6%29%2BC_1%3D%5Cfrac%7Bx%5E%7B12%7D%20%7D%7B2%7D-2x%5E6%2B2%2BC_1%3D%5Cfrac%7Bx%5E%7B12%7D%20%7D%7B2%7D-2x%5E6%2BC)
where ![C=2+C_1](https://tex.z-dn.net/?f=C%3D2%2BC_1)
Without using substitution:
∫
= ∫
= ![\frac{6x^{12} }{12}-\frac{12x^6}{6}+C=\frac{x^{12} }{2}-2x^6+C](https://tex.z-dn.net/?f=%5Cfrac%7B6x%5E%7B12%7D%20%7D%7B12%7D-%5Cfrac%7B12x%5E6%7D%7B6%7D%2BC%3D%5Cfrac%7Bx%5E%7B12%7D%20%7D%7B2%7D-2x%5E6%2BC)
So, same answer is obtained in both the cases.
Answer:
He should make the hypotenuse 5 in
Step-by-step explanation:
the pythagorean heoreom states that a^2+b^2=c^2. c is the hypotenuse. You already have a and b, 3 and 4. Plug it in
3^2+4^2=c^2
9+16=c^2
25=c^2
square root each side to isolate
c=5