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NARA [144]
2 years ago
15

Which of the following statements describes an angle bisector of angle ABC? heagarty maths

Mathematics
1 answer:
Andrej [43]2 years ago
4 0

Answer:

Given an angle ABC, it is possible to construct a line BF that divides the angle into two equal parts using only a straightedge and compass. Such a line is called an angle bisector. ... If we then construct the line BF, it will divide the original angle ABC into two equal angles.

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Write a standard is the number seventy-two and seventeen thousandths then round the nearest hundredths
Natali5045456 [20]

Answer:

72.017

Step-by-step explanation:

Rounded to the nearest hundredths: 72.02

7 0
2 years ago
Read 2 more answers
. Which function's graph has a vertex at (-1, -3) and contains the point (2, 15)?<br> HELPPP PLEASEE
AlexFokin [52]

Answer:

2(x+1)²-3

Step-by-step explanation:

Using the vertex formula we can write the follwoing equation

y=a(x+1)²-3

We need to solve for a

we know that when x= 2 y = 15

So we can write

15=a(2+1)²-3

15=9a-3

18=9a

a=2

Which means our final equation is

2(x+1)²-3

in standard form (just in case)

2x²+4x-1

5 0
3 years ago
A football team lost 5 yards on each of 3 plays. Explain how you could use a number line to find the team's change in field posi
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Answer:

Step-by-step explanation:

5- 10- 15-

3 0
3 years ago
Simplify, 3xsqrt-75a
Naily [24]
Here you go! Hopefully it was correct as i used an app called Cymath.Please mark me as Brainliest!

7 0
3 years ago
A heavy rope, 50 ft long, weighs 0.6 lb/ft and hangs over the edge of a building 120 ft high. Approximate the required work by a
Anastasy [175]

Answer:

Exercise (a)

The work done in pulling the rope to the top of the building is 750 lb·ft

Exercise (b)

The work done in pulling half the rope to the top of the building is 562.5 lb·ft

Step-by-step explanation:

Exercise (a)

The given parameters of the rope are;

The length of the rope = 50 ft.

The weight of the rope = 0.6 lb/ft.

The height of the building = 120 ft.

We have;

The work done in pulling a piece of the upper portion, ΔW₁ is given as follows;

ΔW₁ = 0.6Δx·x

The work done for the second half, ΔW₂, is given as follows;

ΔW₂ = 0.6Δx·x + 25×0.6 × 25 =  0.6Δx·x + 375

The total work done, W = W₁ + W₂ = 0.6Δx·x + 0.6Δx·x + 375

∴ We have;

W = 2 \times \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= 2 \times \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 750

The work done in pulling the rope to the top of the building, W = 750 lb·ft

Exercise (b)

The work done in pulling half the rope is given by W₂ as follows;

W_2 =  \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 562.5

The work done in pulling half the rope, W₂ = 562.5 lb·ft

6 0
2 years ago
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