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3 years ago

Name the point of concurrency of the angle bisectors

Mathematics

1 answer:

Licemer1 [7]3 years ago

3 0

Answer:

The correct option is C.

Step-by-step explanation:

We have to find the point of concurrency of the angle bisectors. It means we have to find the intersection points of all bisectors.

Let the vertices of triangle are PQR.

From the given figure it is notices that the line PC, QC and RC bisects the angles P, Q and R respectively.

The lines PC,QC are RC are bisectors of the triangle and they are intersecting each other at point C.

Therefore point C is point of concurrency of the angle bisectors and option C is correct.

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A rectangle is 8 inches long and 4 inches wide. A similar rectangle is 16 inches long. What is the width of the second rectangle

Shkiper50 [21]

8 × 2 = 16 so..... 4 × 2 = 8

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3 years ago

Graph the line with slope 1/2 passing through the point (-4,4).

Yakvenalex [24]

Answer:

Step-by-step explanation:

------------

It's a straight line, so you need 1 additional point.

---

Plot (-4,4)

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3 years ago

Martin throws a ball straight up in the air. the equation h(t) = -16t^2 + 40t + 5 gives the height of the ball, in feet, t secon

Lisa [10]

Answer:

2.62 seconds

Step-by-step explanation:

Let

t ----> the time in seconds

h(t) ----> he height of the ball, in feet

we have

$h(t)=-16t^2+40t+5$

we know that

When the ball hits the ground, the height is equal to zero

$-16t^2+40t+5=0$

The formula to solve a quadratic equation of the form

$at^{2} +bt+c=0$

is equal to

$t=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-16t^2+40t+5=0$

$a=-16\\b=40\\c=5$

substitute in the formula

$t=\frac{-40\pm\sqrt{40^{2}-4(-16)(5)}} {2(-16)}$

$t=\frac{-40\pm\sqrt{1,920}} {-32}$

$t=\frac{-40+\sqrt{1,920}} {-32}=-0.12\ sec$

$t=\frac{-40-\sqrt{1,920}} {-32}=2.62\ sec$

therefore

The solution is t=2.62 seconds

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Answer:

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