Part A:
We have the following expression:
5x2b2 + 14xb2 - 3b2
Rewriting we have:
b2 (5x2 + 14x - 3)
b2 (5x-1) (x + 3)
Part B:
We have the following expression:
x2 - 18x + 81
Rewriting we have:
(x-9) (x-9)
(x-9) ^ 2
Part C:
We have the following expression:
x2 - 121
Rewriting we have:
(x-11) (x + 11)
Ax + By + C = 0
A. 3x + 5y + 45 = 0
3x + 5(0) + 45 = 0
3x = -45
x = -45/3......x int.....-C/A
x = -15
B. To find the x intercept, u must sub in 0 for y and solve for x
C. This is not true for a horizontal line because u can't sub in 0 for x because a horizontal line never crosses the x axis and therefore, does not have an x axis.
D. y int is -C/B....u find this by subbing in 0 for x and solving for y
3x + 5y + 45 = 0
3(0) + 5y + 45 = 0
5y = -45
y = -45/5......y int = -C/B
y = -9
Answer:
The measures of angles of triangle MNP are
Step-by-step explanation:
<u><em>The picture of the question in the attached figure</em></u>
step 1
Find the measure of arcs AB, BC and AC
we know that
The inscribed angle is half that of the arc it comprises.
so
![\gamma=\frac{1}{2}[arc\ AB] ----> arc\ AB=2\gamma\\\alpha=\frac{1}{2}[arc\ BC] ----> arc\ BC=2\alpha\\\beta=\frac{1}{2}[arc\ AC] ----> arc\ AC=2\beta](https://tex.z-dn.net/?f=%5Cgamma%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20AB%5D%20----%3E%20arc%5C%20AB%3D2%5Cgamma%5C%5C%5Calpha%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20BC%5D%20----%3E%20arc%5C%20BC%3D2%5Calpha%5C%5C%5Cbeta%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20AC%5D%20----%3E%20arc%5C%20AC%3D2%5Cbeta)
step 2
Find the measure of angle M
we know that
The measurement of the outer angle is the semi-difference of the arcs it encompasses.
![M=\frac{1}{2}[arc\ AB+arc\ BC-arc\ AC]](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20AB%2Barc%5C%20BC-arc%5C%20AC%5D)
substitute
![M=\frac{1}{2}[2\gamma+2\alpha-2\beta]\\M=[\gamma+\alpha-\beta]](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B1%7D%7B2%7D%5B2%5Cgamma%2B2%5Calpha-2%5Cbeta%5D%5C%5CM%3D%5B%5Cgamma%2B%5Calpha-%5Cbeta%5D)
step 3
Find the measure of angle N
we know that
The measurement of the outer angle is the semi-difference of the arcs it encompasses.
![N=\frac{1}{2}[arc\ AC+arc\ BC-arc\ AB]](https://tex.z-dn.net/?f=N%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20AC%2Barc%5C%20BC-arc%5C%20AB%5D)
substitute
![N=\frac{1}{2}[2\beta+2\alpha-2\gamma]\\N=[\beta+\alpha-\gamma]](https://tex.z-dn.net/?f=N%3D%5Cfrac%7B1%7D%7B2%7D%5B2%5Cbeta%2B2%5Calpha-2%5Cgamma%5D%5C%5CN%3D%5B%5Cbeta%2B%5Calpha-%5Cgamma%5D)
step 4
Find the measure of angle P
we know that
The measurement of the outer angle is the semi-difference of the arcs it encompasses.
![P=\frac{1}{2}[arc\ AC+arc\ AB-arc\ BC]](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20AC%2Barc%5C%20AB-arc%5C%20BC%5D)
substitute
![P=\frac{1}{2}[2\beta+2\gamma-2\alpha]\\P=[\beta+\gamma-\alpha]](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B1%7D%7B2%7D%5B2%5Cbeta%2B2%5Cgamma-2%5Calpha%5D%5C%5CP%3D%5B%5Cbeta%2B%5Cgamma-%5Calpha%5D)
Step-by-step explanation:
I got about 9 hours of sleep last night that is 9/24 of the day or 37.5% of the day of 0.375 of the day sleeping
thanks for asking
Tape dispenser the others are too heavy
