In order for the triangle to be isosceles, we have to set two lengths of the triangle equal to each other.
Let's take the lengths 5x-12 and x+20 and set them equal to each other.
5x - 12 = x + 20
Combine like terms by moving them over to their respective sides.
Subtract x from both sides of the equation.
4x - 12 = 20
Add 12 to both sides of the equation.
4x = 32
Divide both sides by 4.
x = 8
Check your answer by substituting.
5x - 12 = x + 20
5(8) - 12 = 8 + 20
40 - 12 = 28
28 = 28
Solution: x = 8
Answer:
x=180-x
2x=180
x=90
Step-by-step explanation:
Answer:
7.06 x 10^(-7) ft 3
Step-by-step explanation:
We have the formula to calculate the volume of an octagonal Pyraamid as following:
<em>+) Volume of octagonal pyramid = 1/3 * Area of the base * Height</em>
As given, the base of the pyramid is an octagon with area equal to 15mm2
=> Area of the base = 15 mm2
The height of the pyramid is the length of the line segment which is perpendicular to the base - which is the red line.
=> Height = 4mm
So we have:
<em>Volume of octagonal pyramid = 1/3 * Area of the base * Height</em>
<em>= 1/3 * 15 * 4 = 20 mm3</em>
<em />
As: 1 mm3 = 3.53 x 10^(-8) ft 3
=> 20 mm3 = 7.06x10^(-7) ft 3
So the volume of the pyramid is : 7.06 x 10^(-7) ft 3
we can see that the center is (-3, 3) and the radius is 9 units.
<h3 /><h3>
How to find the center and radius of the circle?</h3>
The general circle equation, for a circle with a center (a, b) and radius R is given by:
(x - a)^2 + (y - b)^2 = R^2
Here we have the equation:
x^2 + y^2 + 6x = 6y + 63
Let's complete squares:
x^2 + y^2 + 6x - 6y = 63
(x^2 + 6x) + (y^2 - 6y) = 63
(x^2 + 2*3x) + (y^2 - 2*3y) = 63
Now we can add and subtract 9, (two times) so we get:
(x^2 + 2*3x + 9) - 9 + (x^2 - 2*3x + 9) - 9 = 63
(x + 3)^2 + (y - 3)^2 = 63 + 9 + 9 = 81 = 9^2
(x + 3)^2 + (y - 3)^2 = 9^2
Comparing with the general circle equation, we can see that the center is (-3, 3) and the radius is 9 units.
If you want to learn more about circles:
brainly.com/question/1559324
#SPJ1
650 * 3456 = 2,246,400 is the answer.
