Answer:
x = -6/5
y = -8/5
Step-by-step explanation:
2x + y = -4
Y = 3x + 2
Substitute the second equation into the first equation
2x + 3x+2 = -4
Combine like terms
5x +2 = -4
Subtract 2 from each side
5x+2-2=-4-2
5x = -6
Divide by 5
5x/5 = -6/5
x = -6/5
Now we need to find y
y = 3x+2
y = 3(-6/5)+2
y = -18/5 + 10/5
y = -8/5
2s+14=12
2s=-14+12
2s=-2
s=-2÷2
s=-1
Answer:
21
Step-by-step explanation:
You do the parentheses first, so 27/9 = 3
10-3 = 7
3 * 7 = 21
Please, use parentheses to enclose each fraction:
y=3/4X+5 should be written as <span>y=(3/4)X+5
Let's eliminate the fraction 3/4 by multiplying the above equation through by 4:
4[y] = 4[(3/4)x + 5]
Then 4y = 3x + 20
(no fraction here)
Let 's now solve the system
4y=3x + 20
4x-3y=-1
We are to solve this system using subtraction. To accomplish this, multiply the first equation by 3 and the second equation by 4. Here's what happens:
12y = 9x + 60 (first equation)
16x-12y = -4, or -12y = -4 - 16x (second equation)
Then we have
12y = 9x + 60
-12y =-16x - 4
If we add here, 12y-12y becomes zero and we then have 0 = -7x + 56.
Solving this for x: 7x = 56; x=8
We were given equations
</span><span>y=3/4X+5
4x-3y=-1
We can subst. x=8 into either of these eqn's to find y. Let's try the first one:
y = (3/4)(8)+5 = 6+5=11
Then x=8 and y=11.
You should check this result. Subst. x=8 and y=11 into the second given equation. Is this equation now true?</span>
Given the figure of a regular pyramid
The base of the pyramid is a hexagon with a side length = 6
The lateral area is 6 times the area of one of the side triangles
So, the side triangle has a base = 6
The height will be:
![\begin{gathered} h^2=6^2+(\frac{\sqrt[]{3}}{2}\cdot6)^2=36+27=63 \\ h=\sqrt[]{63} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20h%5E2%3D6%5E2%2B%28%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B2%7D%5Ccdot6%29%5E2%3D36%2B27%3D63%20%5C%5C%20h%3D%5Csqrt%5B%5D%7B63%7D%20%5Cend%7Bgathered%7D)
so, the lateral area =