Answer:
Step-by-step explanation:
Solve for x=0 and x=1
y(0)=2(3^0)=2(1)=2 so (0,2) is a point.
y(1)=2(3^1)=2(3)=6 so (1,6) is a point.
Only graph B has those two points.
Answer:
x = 10
Step-by-step explanation:
9x - 40 = 3x + 20
<u>9</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>x</u> - 40 = <u>3x - 3x</u> + 20
6x - 40 = 20
6x <u>-</u><u> </u><u>40</u><u> </u><u>+</u><u> </u><u>40</u> = <u>20</u><u> </u><u>+</u><u> </u><u>40</u>
6x = 60
<u>6x</u><u> </u><u>/</u><u> </u><u>6</u> = <u>60</u><u> </u><u>/</u><u> </u><u>6</u>
x = 10
Now plug the x value in the equation to make the statement true that A is parallel to B.
9x - 40
<u>9</u><u>(</u><u>10</u><u>)</u> - 40
<u>90</u><u> </u><u>-</u><u> </u><u>40</u>
50
3x + 20
<u>3</u><u>(</u><u>10</u><u>)</u> + 20
<u>30</u><u> </u><u>+</u><u> </u><u>20</u>
50
Therefore, x = 10 making the statement true that A is parallel to B. Hope this helps and stay safe, happy, and healthy, thank you :) !!
Answer:
1. x = 15, y = 8
2. m∠MSN = 90°
3. x = 20
4. y = 9
5. x = 8, m∠PQS = 24, m∠SQR = 66
6. y = 15, m∠RPT = 55, m∠TPW = 35
Step-by-step explanation:
1. to solve for y
9y + 18 = 90
9y + 18 - 18 = 90 - 18
9y = 72
9y/9 = 72/9
y = 8
to solve for x
5x + x = 90
6x = 90
6x/6 = 90/6
x = 15
2. m∠MSN = 90°, because ∠MSN is a right angle, which is equal to 90
3. To find x
3x + 10 + x = 90
3x + x = 4x
4x + 10 = 90
4x + 10 - 10 = 90 - 10
4x = 80
4x/4 = 80/4
x = 20
4. To find y
7y - 3 + 3y + 3 = 90
7y + 3y = 10
-3 + 3 = 0
10y = 90
10y/10 = 90/10
y = 9
5. To find x
3x + 8x + 2 = 90
3x + 8x = 11x
11x + 2 = 90
11x + 2 - 2 = 90 -2
11x = 88
11x/11 = 88/11
x = 8
To find ∠PQS
∠PQS = 3x
x = 8
so 3(x) = 3(8)
3(8) = 24
so ∠PQS = 24
To find ∠SQR
∠SQR = 8x + 2
x = 8
so ∠SQR = 8(x) + 2 = 8(8) + 2
8(8) + 2
= 64 + 2
= 66
6. To find y
4y - 5 + 2y + 5 = 90
4y + 2y = 6y
-5 + 5 = 0
6y = 90
6y/6 = 90/6
y = 15
To find ∠RPT
∠RPT = 4y - 5
y = 15
so ∠RPT = 4y - 5 = 4(15) - 5
4(15) - 5
= 60 - 5
= 55
To find ∠TPW
∠TPW = 2y + 5
y = 15
so ∠TPW = 2y + 5 = 2(15) + 5
2(15) + 5
= 30 + 5
= 35
Answer:
c
Step-by-step explanation:

Answer:
We are given that triangle PQR is an isosceles triangle in which PQ = PR .
Since the base angles of an isosceles triangle are equal,
angle PQR = angle PRQ
Also, And we are given that
angle MRQ = angle NQR
In ΔQNR and ΔRMQ
∠NQR=∠MRQ (given)
QR = QR (common)
-triangles QNR is congruent to triangles RMQ - ASA - angle side angle