Answer:
n= -10
Step-by-step explanation:
-66=7n+4
-7n=4+66
-7n=70
n=70/-7
n= -10
<u>Answer:</u>
A) slope of equation 3x+4y-12=0 is m=
and y-intercept is 3
B) slope of equation 2x+y=1 is m=-2 and y-intercept is 1.
C) slope of equation 2x-y=0 is m=2 and y-intercept is 0.
<u>Solution:</u>
The slope - intercept form equation of line is given as
y=mx+c --- eqn (1)
Where m is the slope of the line. The coefficient of “x” is the value of slope of the line.
c is the y – intercept which is the value of y at the point where the line crosses the y-axis
a) Given that
3x+4y-12=0 --- eqn (2)
On converting equation (2) in slope intercept form, we get
4y=-3x+12
![y=-\frac{3}{4} x+\frac{12}{4}](https://tex.z-dn.net/?f=y%3D-%5Cfrac%7B3%7D%7B4%7D%20x%2B%5Cfrac%7B12%7D%7B4%7D)
![y=-\frac{3}{4} x+3 --- eqn 3](https://tex.z-dn.net/?f=y%3D-%5Cfrac%7B3%7D%7B4%7D%20x%2B3%20---%20eqn%203)
On comparing equation (3) and (1) we get slope of equation (2) is
and y-intercept is 3.
b) Given that
2x+y=1 …(4)
On converting equation (2) in slope intercept form, we get
y=-2x+1 …(5)
On comparing equation (5) and (1) we get slope of equation (4) is m=-2 and y-intercept is 1.
C)Given that
2x-y=0 …(6)
On converting equation (6) in slope intercept form, we get
y=2x …(7)
On comparing equation (7) and (1) we get slope of equation (6) is m=2 and y-intercept is 0.
-2(2b-1)+4b
step 1- distribute: -4b-2+4b
step 2- add: -4b+4b=0
the answer is -2
Answer:
3
Step-by-step explanation:
![x+3\neq 0\\x\neq -3\\\frac{x^2}{x+3} =\frac{9}{x+3} \\x^2=9\\x=3](https://tex.z-dn.net/?f=x%2B3%5Cneq%200%5C%5Cx%5Cneq%20-3%5C%5C%5Cfrac%7Bx%5E2%7D%7Bx%2B3%7D%20%3D%5Cfrac%7B9%7D%7Bx%2B3%7D%20%5C%5Cx%5E2%3D9%5C%5Cx%3D3)
Answer:
The area of the base of the pyramid is 109.2 mm.
Step-by-step explanation:
The area of the base of a hexagonal pyramid is given by the area of a hexagon:
Where:
P: is the perimeter
a: is the apothem
We need to find the perimeter and the apothem.
The perimeter is equal to:
![P = 6*s](https://tex.z-dn.net/?f=%20P%20%3D%206%2As%20)
Where:
s: is the side of the pyramid
And the apothem is:
![a = \frac{\sqrt{3}}{2}*s](https://tex.z-dn.net/?f=%20a%20%3D%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%2As%20)
So, to calculate the apothem and the perimeter we need to calculate the side of the pyramid. We can find it from the volume of the pyramid:
Where:
h: is the height = 4 mm
V: is the volume = 144 mm³
Then, the side is:
![s = \sqrt{\frac{2V}{\sqrt{3}*h}} = \sqrt{\frac{2*144}{\sqrt{3}*4}} = 6.5 mm](https://tex.z-dn.net/?f=%20s%20%3D%20%5Csqrt%7B%5Cfrac%7B2V%7D%7B%5Csqrt%7B3%7D%2Ah%7D%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B2%2A144%7D%7B%5Csqrt%7B3%7D%2A4%7D%7D%20%3D%206.5%20mm%20)
Now, we can find the perimeter and the apothem.
![P = 6*s = 6*6.5 = 39 mm](https://tex.z-dn.net/?f=%20P%20%3D%206%2As%20%3D%206%2A6.5%20%3D%2039%20mm%20)
Finally, the area is:
Therefore, the area of the base of the pyramid is around 109 mm.
I hope it helps you!