The circumference of a circle is 35 pi. What is the diameter?
2 answers:
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<span>4. The x-coordinate of the solution is 9.
</span><span>5. The y-coordinate of the solution is −8 .
</span><span>6.The ordered pair that is the solution to the system lies in Quadrant IV .</span><span>
Proof:
Solve the following system:
{4 x + 3 y = 12 | (equation 1)
{2 x + 3 y = -6 | (equation 2)
Subtract 1/2 × (equation 1) from equation 2:
{4 x + 3 y = 12 | (equation 1)
{0 x+(3 y)/2 = -12 | (equation 2)
Multiply equation 2 by 2/3:
{4 x + 3 y = 12 | (equation 1)
{0 x+y = -8 | (equation 2)
Subtract 3 × (equation 2) from equation 1:
{4 x+0 y = 36 | (equation 1)
{0 x+y = -8 | (equation 2)
Divide equation 1 by 4:
{x+0 y = 9 | (equation 1)
{0 x+y = -8 | (equation 2)
Collect results:
Answer: {x = 9 , y = -8
</span>
</span><span>5. The y-coordinate of the solution is −8 .
</span><span>6.The ordered pair that is the solution to the system lies in Quadrant IV .</span><span>
Proof:
Solve the following system:
{4 x + 3 y = 12 | (equation 1)
{2 x + 3 y = -6 | (equation 2)
Subtract 1/2 × (equation 1) from equation 2:
{4 x + 3 y = 12 | (equation 1)
{0 x+(3 y)/2 = -12 | (equation 2)
Multiply equation 2 by 2/3:
{4 x + 3 y = 12 | (equation 1)
{0 x+y = -8 | (equation 2)
Subtract 3 × (equation 2) from equation 1:
{4 x+0 y = 36 | (equation 1)
{0 x+y = -8 | (equation 2)
Divide equation 1 by 4:
{x+0 y = 9 | (equation 1)
{0 x+y = -8 | (equation 2)
Collect results:
Answer: {x = 9 , y = -8
</span>
Answer:
B. The student incorrectly calculated the scale factor to be –2
Step-by-step explanation:
Given that :
After a dilation with a center of (0, 0), a point was mapped as (4, –6) → (12, y).
The student determined y to be -2
If a figure dilated with a center of (0, 0) and scale factor k, then
(x , y) → (kx , ky)
(4, -6) → (12, y)
k = 3
Thus; the scale factor is 3
Now; the y-coordinate can now be calculated as;
ky = (3 × -6)
ky = -18
Therefore; the value of y = -18 and the student incorrectly calculated the scale factor to be -2.