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BigorU [14]
2 years ago
7

Given the figure shown at the right ABC is a triangle, what would be the measure of the unknown exterior angle (labeled x in the

diagram)?

Mathematics
1 answer:
Andre45 [30]2 years ago
4 0

Answer:

d. 112°

Step-by-step explanation:

m<A = 64° (given)

m<ABC = 180° - 132° (linear pair)

m<ABC = 48°

According to the exterior angle theorem of a triangle, the exterior angle of a ∆ is equal to the opposite interior angles of the ∆.

48° and 64° are interior angles of ∆ABC that are opposite to the exterior angle, x.

Therefore,

x = 48 + 64

x = 112°

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Read 2 more answers
Let T:P3→P3 be the linear transformation such that T(â’2x2)=3x2+4x, T(0.5x+3)=â’3x2+3xâ’4, and T(2x2â’1)=â’3x+4. Find T(1), T(
balu736 [363]

It looks like we're told that

T(-2x^2)=3x^2+4x

T\left(\dfrac12x+3\right)=-3x^2+3x-4

T(2x^2-1)=-3x+4

We use the fact that T is linear to find T(1),T(x),T(x^2). First, we notice that

T(-2x^2)+T(2x^2-1)=T(-2x^2+2x^2-1)=T(-1)=-T(1)

We also have

T\left(\dfrac12x+3\right)=T\left(\dfrac12x\right)+T(3)=\dfrac12T(x)+3T(1)

T(2x^2-1)=T(2x^2)-T(1)=2T(x^2)-T(1)

So once we find T(1), we can determine T(x) and T(x^2). We have

T(1)=-\left(T(-2x^2)+T(2x^2-1)\right)\implies T(1)=-3x^2-x-4

and using this we find

T(x)=12x^2+12x+16

T(x^2)=-\dfrac32x^2-2x

Then

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T(ax^2+bx+c)=-\dfrac32\left(a-8b+2c\right)x^2-(2a-12b+c)x+16b-4c

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