Answer:
The answer is 82 unit total
Step-by-step explanation:
You can take the entire shape and break it down to normal shapes such as squares and triangles. Then you can use the formula of these shapes to find the area. Last you add all the unit together and that's the answer
Answer with Step-by-step explanation:
We are given that
u+ v and u-v are orthogonal
We have to prove that u and v must have the same length.
When two vector a and b are orthogonal then
![a\cdot b=0](https://tex.z-dn.net/?f=a%5Ccdot%20b%3D0)
By using the property
![(u+v)\cdot (u-v)=0](https://tex.z-dn.net/?f=%28u%2Bv%29%5Ccdot%20%28u-v%29%3D0)
We know that
![(a+b)\cdot (a-b)=\mid a\mid^2-\mid b\mid^2](https://tex.z-dn.net/?f=%28a%2Bb%29%5Ccdot%20%28a-b%29%3D%5Cmid%20a%5Cmid%5E2-%5Cmid%20b%5Cmid%5E2)
![\mid u\mid ^2-\mid v\mid^2=0](https://tex.z-dn.net/?f=%5Cmid%20u%5Cmid%20%5E2-%5Cmid%20v%5Cmid%5E2%3D0)
![\mid u\mid^2=\mid v\mid^2](https://tex.z-dn.net/?f=%5Cmid%20u%5Cmid%5E2%3D%5Cmid%20v%5Cmid%5E2)
Magnitude is always positive
When power of base on both sides are equal then base will be equal
Therefore,
![\mid u\mid=\mid v\mid](https://tex.z-dn.net/?f=%5Cmid%20u%5Cmid%3D%5Cmid%20v%5Cmid)
Hence, the length of vectors u and v must have the same length.
Answer:
Option 3
y + 16 = 8(x + 2)
Step-by-step explanation:
<h3>
<u>Given</u>;</h3>
- x = 2, 4, 6, 8
- y = 16, 32, 48, 64
Now, Put the values in equation 1
y – 16 = 8(x – 2)
y – 16 = 8(x – 2)
16 – 16 = 8(2 – 2)
0 = 8
Here, L.H.S ≠ R.H.S
So, Option 1 is incorrect
Now, Put the values in equation 2
y – 16 = 8x – 2
y – 16 = 8x – 2
16 – 16 = 8(2) – 2
0 = 14
Here, L.H.S ≠ R.H.S
So, Option 1 is incorrect
Now, Put the values in equation 3
y + 16 = 8(x + 2)
y + 16 = 8(x + 2)
16 + 16 = 8(2 + 2)
32 = 8(4)
32 = 32
Here, L.H.S = R.H.S
Check other values of x and y
y + 16 = 8(x + 2)
32 + 16 = 8(4 + 2)
48 = 8(6)
48 = 48
Here, L.H.S = R.H.S
y + 16 = 8(x + 2)
48 + 16 = 8(6 + 2)
64 = 8(8)
64 = 64
Here, L.H.S = R.H.S
y + 16 = 8(x + 2)
64 + 16 = 8(8 + 2)
80 = 8(10)
80 = 80
Here, L.H.S = R.H.S
Thus, The equation 3 in point-slope from gives the plant's height at any time.
<u>-TheUnknownScientist 72</u>
X=2 and you go up on the y-intercept 2 times and move 1 right.
a. This information is given to you.
b. We want to find
![\mathrm{Pr}\{X > 8.9\}](https://tex.z-dn.net/?f=%5Cmathrm%7BPr%7D%5C%7BX%20%3E%208.9%5C%7D)
so we first transform
to the standard normal random variable
with mean 0 and s.d. 1 using
![X = \mu + \sigma Z](https://tex.z-dn.net/?f=X%20%3D%20%5Cmu%20%2B%20%5Csigma%20Z)
where
are the mean/s.d. of
. Now,
![\mathrm{Pr}\left\{\dfrac{X - 10.5}2 > \dfrac{8.9 - 10.5}2\right\} = \mathrm{Pr}\{Z > -0.8\} \\\\~~~~~~~~= 1 - \mathrm{Pr}\{Z\le-0.8\} \\\\ ~~~~~~~~ = 1 - \Phi(-0.8) \approx \boxed{0.7881}](https://tex.z-dn.net/?f=%5Cmathrm%7BPr%7D%5Cleft%5C%7B%5Cdfrac%7BX%20-%2010.5%7D2%20%3E%20%5Cdfrac%7B8.9%20-%2010.5%7D2%5Cright%5C%7D%20%3D%20%5Cmathrm%7BPr%7D%5C%7BZ%20%3E%20-0.8%5C%7D%20%5C%5C%5C%5C~~~~~~~~%3D%201%20-%20%5Cmathrm%7BPr%7D%5C%7BZ%5Cle-0.8%5C%7D%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%201%20-%20%5CPhi%28-0.8%29%20%5Capprox%20%5Cboxed%7B0.7881%7D)
where
is the CDF for
.
c. The 76th percentile is the value of
such that
![\mathrm{Pr}\{X \le x_{76}\} = 0.76](https://tex.z-dn.net/?f=%5Cmathrm%7BPr%7D%5C%7BX%20%5Cle%20x_%7B76%7D%5C%7D%20%3D%200.76)
Transform
to
and apply the inverse CDF of
.
![\mathrm{Pr}\left\{Z \le \dfrac{x_{76} - 10.5}2\right\} = 0.76](https://tex.z-dn.net/?f=%5Cmathrm%7BPr%7D%5Cleft%5C%7BZ%20%5Cle%20%5Cdfrac%7Bx_%7B76%7D%20-%2010.5%7D2%5Cright%5C%7D%20%3D%200.76)
![\dfrac{x_{76} - 10.5}2 = \Phi^{-1}(0.76)](https://tex.z-dn.net/?f=%5Cdfrac%7Bx_%7B76%7D%20-%2010.5%7D2%20%3D%20%5CPhi%5E%7B-1%7D%280.76%29)
![\dfrac{x_{76} - 10.5}2 \approx 0.7063](https://tex.z-dn.net/?f=%5Cdfrac%7Bx_%7B76%7D%20-%2010.5%7D2%20%5Capprox%200.7063)
![x_{76} - 10.5 \approx 1.4126](https://tex.z-dn.net/?f=x_%7B76%7D%20-%2010.5%20%5Capprox%201.4126)
![x_{76} \approx \boxed{11.9126}](https://tex.z-dn.net/?f=x_%7B76%7D%20%5Capprox%20%5Cboxed%7B11.9126%7D)