X-intercept => 2x = 6 => x = 3
x-intercept is 3.
y-intercept => 3/2y = 6 => y = 4
y-intercept is 4.
z-intercept => 3z = 6 => z = 2
z-intercept is 2.
Volume of a quadrangular pyramid=(1/3)bh
b=base
h=height
b=area of the base=area of a square=8.4 ft * 8.4 ft=70.56 ft²
Pythagoras theorem:
hypotenuse²=leg₁² + leg₂²
data:
hypotenuse=9.6 ft
leg₁=height=h
leg₂=8.4 ft /2=4.2 ft
(9.6 ft)²= h² + (4.2 ft)²
92.16 ft²=h²+17.64 ft²
h²=92.16 ft²-17.64 ft²
h²=74.52 ft²
h=√(74.52 ft²)=8.63 ft.
Volume of this quadrangular pyramid=(1/3)(70.56 ft²)(8.63 ft)=202.9 ft³≈202.3 ft³
Answer: ≈202.3 ft³
Answer:
35 lol
Step-by-step explanation:
a = b*h/2
a = 10*7/2
a = 70/2
a = 35 lol
Answer:
its B :)
Step-by-step explanation:
I hope this helps ^^
We have to solve this equation:
Third degree polynomials like this one are not easily solved, but this one has a root at x = 0. The let us factorize this polynomial as x times a second degree polynomial:
Now we can find the roots of the quadratic polynomial as:
![\begin{gathered} x=\frac{-(-6)\pm\sqrt[]{(-6)^2-4\cdot1\cdot6}}{2\cdot1} \\ x=\frac{6\pm\sqrt[]{36-24}}{2} \\ x=\frac{6\pm\sqrt[]{12}}{2} \\ x=\frac{6\pm\sqrt[]{4\cdot3}}{2} \\ x=\frac{6\pm2\sqrt[]{3}}{2} \\ x=3\pm\sqrt[]{3} \\ x_1=3-\sqrt[]{3} \\ x_2=3+\sqrt[]{3} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%3D%5Cfrac%7B-%28-6%29%5Cpm%5Csqrt%5B%5D%7B%28-6%29%5E2-4%5Ccdot1%5Ccdot6%7D%7D%7B2%5Ccdot1%7D%20%5C%5C%20x%3D%5Cfrac%7B6%5Cpm%5Csqrt%5B%5D%7B36-24%7D%7D%7B2%7D%20%5C%5C%20x%3D%5Cfrac%7B6%5Cpm%5Csqrt%5B%5D%7B12%7D%7D%7B2%7D%20%5C%5C%20x%3D%5Cfrac%7B6%5Cpm%5Csqrt%5B%5D%7B4%5Ccdot3%7D%7D%7B2%7D%20%5C%5C%20x%3D%5Cfrac%7B6%5Cpm2%5Csqrt%5B%5D%7B3%7D%7D%7B2%7D%20%5C%5C%20x%3D3%5Cpm%5Csqrt%5B%5D%7B3%7D%20%5C%5C%20x_1%3D3-%5Csqrt%5B%5D%7B3%7D%20%5C%5C%20x_2%3D3%2B%5Csqrt%5B%5D%7B3%7D%20%5Cend%7Bgathered%7D)
Then, the solutions to the equation are:
x = 0
x = 3 - √3
x = 3 + √3
