This question gives you information to make a system of equations, which you then use substitution to solve.
Answer with explanation:
The given function is
y=f(x)=x³ -5
Function with Translation
![1.\rightarrow f(\frac{x}{5})=[\frac{x}{5}]^3-5\\\\y_{1}=\frac{x^3}{125}-5\\\\2..\rightarrow f(x-5)=(x-5)^3-5\\\\y_{2}=x^3-5^3-3x^2\times 5+3 x \times 5^2-5\\\\y_{2}=x^3-15 x^2+75 x-130 \\\\3.\rightarrow f(x)+10=x^3-5+10\\\\y_{3}=x^3+5\\\\4.\rightarrow f(5 x-2)=(5 x-2)^3-5\\\\y_{4}=(5 x)^3-(2)^3-3 \times (5 x)^2 \times 2+3 \times (5 x) \times (2)^2-5\\\\y_{4}=125 x^3-8-150 x^2+60 x-5\\\\y_{4}=125 x^3-150 x^2+60 x-13](https://tex.z-dn.net/?f=1.%5Crightarrow%20f%28%5Cfrac%7Bx%7D%7B5%7D%29%3D%5B%5Cfrac%7Bx%7D%7B5%7D%5D%5E3-5%5C%5C%5C%5Cy_%7B1%7D%3D%5Cfrac%7Bx%5E3%7D%7B125%7D-5%5C%5C%5C%5C2..%5Crightarrow%20f%28x-5%29%3D%28x-5%29%5E3-5%5C%5C%5C%5Cy_%7B2%7D%3Dx%5E3-5%5E3-3x%5E2%5Ctimes%205%2B3%20x%20%5Ctimes%205%5E2-5%5C%5C%5C%5Cy_%7B2%7D%3Dx%5E3-15%20x%5E2%2B75%20x-130%20%5C%5C%5C%5C3.%5Crightarrow%20f%28x%29%2B10%3Dx%5E3-5%2B10%5C%5C%5C%5Cy_%7B3%7D%3Dx%5E3%2B5%5C%5C%5C%5C4.%5Crightarrow%20f%285%20x-2%29%3D%285%20x-2%29%5E3-5%5C%5C%5C%5Cy_%7B4%7D%3D%285%20x%29%5E3-%282%29%5E3-3%20%5Ctimes%20%285%20x%29%5E2%20%5Ctimes%202%2B3%20%5Ctimes%20%285%20x%29%20%5Ctimes%20%282%29%5E2-5%5C%5C%5C%5Cy_%7B4%7D%3D125%20x%5E3-8-150%20x%5E2%2B60%20x-5%5C%5C%5C%5Cy_%7B4%7D%3D125%20x%5E3-150%20x%5E2%2B60%20x-13)
Answer:
x = 34
Step-by-step explanation:
Add together all three angles because together they equal 180 degrees
x + 12 + x + 100 = 180 Combine like terms
2x + 112 = 180
- 112 - 112 Subtract 112 from both sides
2x = 68 Divide both sides by 2
x = 34
Check the picture below.
first off let's check what "h" is in the triangular face.
so, the gazebo has 4 squarish faces, each 12x12, recall 48 ÷ 4 = 12, and it has 4 triangular faces, each with a height of 8 and a base of 12.
notice we're skipping the top and bottom of the cube and the bottom of the pyramid because they're not part of the "surface area".
![\stackrel{\textit{\large Areas}}{\stackrel{\textit{4 triangular faces}}{4\left[\cfrac{1}{2}(12)(8) \right]}~~~~+~~~~\stackrel{\textit{4 squarish faces}}{4[12\cdot 12]}}\implies 192+576\implies 768](https://tex.z-dn.net/?f=%5Cstackrel%7B%5Ctextit%7B%5Clarge%20Areas%7D%7D%7B%5Cstackrel%7B%5Ctextit%7B4%20triangular%20faces%7D%7D%7B4%5Cleft%5B%5Ccfrac%7B1%7D%7B2%7D%2812%29%288%29%20%5Cright%5D%7D~~~~%2B~~~~%5Cstackrel%7B%5Ctextit%7B4%20squarish%20faces%7D%7D%7B4%5B12%5Ccdot%2012%5D%7D%7D%5Cimplies%20192%2B576%5Cimplies%20768)
Answer:
44,39 in
Step-by-step explanation:
To find the space diagonal (whatever it's called in English) you begin looking at the bottom of the box. We want to know the diagonal since the diagonal is one of the sides of the other triangle.
To do this we can start by using trigonometry. (SOH-CAH-TOA)
We need to use Sin t. (Since we wanna know the hypothenuse and we have the opposite.)
So...
Sin(40) = 24 in ÷ <em>h</em>
Then we need to actually be able to calculate this, which we will be able to do if we multiply with h on both sides.
hSin(40) = 24 in
Like that. And now we can divide both sides with Sin(40). So we calculate 24 in devided by Sin(40).
h = 24 in ÷ Sin(40) = 37,34 in
Okay so now we now the diagonal, 37,34 in. Now we can use Pythagorean theorem. (2a+2b=2c and c = square root of (2a+2b))
37,34 in^(2) + 24 in^(2) = 1970,28
And now we take the square root of 1970,28.
Which is about 44,39 in.
<em>If you're not familiar with these concepts I suggest you</em><em> </em><em>search</em><em> </em><em>them</em><em> </em><em>up</em><em>.</em>
