You find the measure of angle BEC, by adding angle B and angle C which is 43+90, and then subtract that from 180 and get 180-(43+90). Then you take the measure of angle BEC (which is the same as AEB), take AEB and A which is 90 degrees and add those together. Then you take that and subtract it from 180, which is 180-(AEB+90)
Answer:
See explanation
Step-by-step explanation:
If
then triangle PXY is isosceles triangle. Angles adjacent to the base XY of an isosceles triangle PXY are congruent, so

and

Angles 1 and 3 are supplementary, so

Angles 2 and 4 are supplementary, so

By substitution property,

Hence,

Consider triangles APX and BPY. In these triangles:
- given;
- given;
- proven,
so
by ASA postulate.
Congruent triangles have congruent corresponding sides, then

Therefore, triangle APB is isosceles triangle (by definition).
Answer:
Let v(t) be the velocity of the car t hours after 2:00 PM. Then
. By the Mean Value Theorem, there is a number c such that
with
. Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly
.
Step-by-step explanation:
The Mean Value Theorem says,
Let be a function that satisfies the following hypotheses:
- f is continuous on the closed interval [a, b].
- f is differentiable on the open interval (a, b).
Then there is a number c in (a, b) such that

Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.
By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.
Let v(t) be the velocity of the car t hours after 2:00 PM. Then
and
(note that 20 minutes is
of an hour), so the average rate of change of v on the interval
is

We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in
at which
.
c is a time time between 2:00 and 2:20 at which the acceleration is
.
Solution :
2a + 2b = 7 ...1)
4a + 3b = 12 ...2)
In equation, 1)
a = (7 - 2b)/2 ...3)
Putting value of a in equation 2) we get :
4 × (7 - 2b)/2 + 3b = 12
2( 7 - 2b ) + 3b = 12
14 - 4b + 3b = 12
b = 2
Putting value of b in 3) we get :
a = ( 7 - 2×2)/2
a = 3/2 = 1.5
Now,
2x - 3y = 16 ...5)
x + 2y = -6 ...6)
x = -6 - 2y
Putting above value of x in eq 5) , we get :
2( -6 - 2y ) - 3y = 16
-12 - 4y - 3y = 16
7y = -28
y = -4
x = -6 - ( 2× -4 )
x = 2
Hence, this is the required solution.