Sam rotated parallelogram ABCD 90° clockwise around the origin. If angle A is 130° and angle B is 50°, what is the degree measurement of angle A'?
Answers
Answer-
The measurement of angle A' will be 130° .
Solution-
Transformations like - rotations, reflections, and translations are isometric. That means that these transformations do not change the size of the figure. If the size and shape of the figure is not changed, then the figures are congruent.
It doesn't matter the order or how much degree of rotation has taken place, the final image will be congruent to the original image.
So, even after the rotation of 90° clockwise around the origin, the measurement of angle A will be 130° and so do all the rest of the angles.
Answer:
B, 130 degrees.
Step-by-step explanation:
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Can someone explain and answer this for me??
Answers
Answer:
h(2)= -17
Step-by-step explanation:
Answer:
h(x) = 31
Step-by-step explanation:
When it says find h (2), you will have to replace x with 2
h(x) = 3x^2 -5
h(x) = 3(2)^2 -5
h(x) = 6^2 - 5
h(x) = 36 - 5
h(x) = 31
Hope this helps =)
Answers
Answer:
-6
Explanation:
(-3) · 2
= -6
[Think of it as -3, twice]
(-3) + (-3)
= -6
Answer:
-3×2 = -6 (you add the negative to the 6 because negative times positive is negative)
Answers
Step-by-step explanation:
A statistic is a percentage of a sample.
A parameter is a percentage of a population.
So 74% is a statistic and 58% is a parameter.
Answers
K(6, 4), L(-6, 4)
Can you lead me through the steps because I'm not getting this at all...
Answers
gradient m = (y2 - y1) / ( x2 - x1)
from the given coordinates, x1 = 6, x2 = -6, y1= 4, y2 = 4
m = (4 - 4) / ( -6 - 6) = 0/12 = 0
to find y- intercept, we solve for c using any of the given coordinates.
4 = 0x + c
c = 4
the equation of the line: y = 0x + 4
Standard form = ax + by = c
0x + y = 4
y = 4
f(x)=-3x^2-5x+1 and point (1,-7)
Answers
f(x)-f(a)= -3x²-5x+1-(-3a²-5a+1)=-3(x²-a²)-5(x-a)=-3(x-a)(x+a)-5(x-a)
=-(x-a)(3(x+a)+5)
=-(x-a)(3x+3a+5)
lim (f(x)-f(a))/(x-a)=- lim (3x+3a+5)=3a+3a+5=-6a-5
if a=1==>-6*1-5=-11
Otherwise
f'(x)=-6x-5
f'(1)=-6-5=-11
at point(1,-7)