Answer:
6.75 or 6 3/4
Step-by-step explanation:
First take the whole number and multiply that by 3.
2*3=6
Now, take the fraction and multiply that times 3.
1/4 as a decimal is 0.25. You can ignore the decimal point right now, but you have to remember it is there later.
So, whithout the decimal point, 25*3= 75.
Now, you add the decimal. The 75 is now 0.75, with the decimal.
Add the whole number and the decimal, and you have your answer!
6+0.75=6.75. As a fraction this is 6 3/4
Answer:
<h2>Yes. It's a right triangle.</h2>
Step-by-step explanation:
If a ≤ b < c are the length of the sides of a right triangle, then
a² + b² = c².
We have:
a =30 ft, b = 40 ft and c = 50 ft.
Check the equality:
L = 30² + 40² = 900 + 1600 = 2500
R = 50² = 2500
L = R
Answer:
The answer A
Step-by-step explanation:
B: You need a side that is longer than 2 equal sides. B is incorrect.
A: Try
9^2 + 12^2 = 15^2 if that works, A will work.
81 + 144 = 225
This works. I used these numbers because they were easier to see.
0.9^2 + 1.2^2 = 1.5^2
0.81 + 1.44 = 2.25
2.25 = 2.25
That is the answer A
But we have to check the other two.
C:
0.9^2 = 0.81
1.2^2 = 1.44
1.8^2 = 3.24 This won't work. 3.24 is just too big.
D:
0.9^2 = 0.81
0.6^2 = 0.36
1.5^2 = 2.25 It's too far away.
Answer:
36π mi
Step-by-step explanation:
Area of a circle = πr²
Diameter = 12 mi. Therefore, radius=12/2 = 6 mi
πr²
π x (6)²
π x 36
Hence, the answer is 36π mi
Answer:
Required solution (a)
(b) 40.
Step-by-step explanation:
Given,
![F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k](https://tex.z-dn.net/?f=F%28x%2Cy%2Cz%29%3Dyz%20%5Cuvec%20i%20%2Bxz%5Cuvec%20j%2B%28xy%2B6z%29%5Cuvex%20k)
(a) Let,
![F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k=f_x \uvec{i} +f_y \uvex{j}+f_z\uvec{k}](https://tex.z-dn.net/?f=F%28x%2Cy%2Cz%29%3Dyz%20%5Cuvec%20i%20%2Bxz%5Cuvec%20j%2B%28xy%2B6z%29%5Cuvex%20k%3Df_x%20%5Cuvec%7Bi%7D%20%2Bf_y%20%5Cuvex%7Bj%7D%2Bf_z%5Cuvec%7Bk%7D)
Then,
![f_x=yz,f_y=xz,f_z=xy+6z](https://tex.z-dn.net/?f=f_x%3Dyz%2Cf_y%3Dxz%2Cf_z%3Dxy%2B6z)
Integrating
we get,
![f(x,y,z)=xyz+g(y,z)](https://tex.z-dn.net/?f=f%28x%2Cy%2Cz%29%3Dxyz%2Bg%28y%2Cz%29)
Differentiate this with respect to y we get,
compairinfg with
of the given function we get,
![g'(y,z)=0\implies g(y,z)=0+h(z)\implies g(y,z)=h(z)](https://tex.z-dn.net/?f=g%27%28y%2Cz%29%3D0%5Cimplies%20g%28y%2Cz%29%3D0%2Bh%28z%29%5Cimplies%20g%28y%2Cz%29%3Dh%28z%29)
Then,
![f(x,y,z)=xyz+h(z)](https://tex.z-dn.net/?f=f%28x%2Cy%2Cz%29%3Dxyz%2Bh%28z%29)
Again differentiate with respect to z we get,
![f_z=xy+h'(z)=xy+6z](https://tex.z-dn.net/?f=f_z%3Dxy%2Bh%27%28z%29%3Dxy%2B6z)
on compairing we get,
(By integrating h'(z)) where C is integration constant. Hence,
![f(x,y,z)=xyz+3z^2+C](https://tex.z-dn.net/?f=f%28x%2Cy%2Cz%29%3Dxyz%2B3z%5E2%2BC)
(b) Next, to find the itegration,
![\int_C \vec{F}.dr=\int_C \nabla f. d\vec{r}=f(5,4,2)-f(2,0,-2)=(52+C)-(12+C)=40](https://tex.z-dn.net/?f=%5Cint_C%20%5Cvec%7BF%7D.dr%3D%5Cint_C%20%5Cnabla%20f.%20d%5Cvec%7Br%7D%3Df%285%2C4%2C2%29-f%282%2C0%2C-2%29%3D%2852%2BC%29-%2812%2BC%29%3D40)