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timurjin [86]
2 years ago
10

An inscribed angle in the diagram is EDG ABF HFA CDE

Mathematics
1 answer:
koban [17]2 years ago
4 0

Answer:

I am pretty sure it is B Angle ABF

Step-by-step explanation:

Because an inscribed angle is a angle formed in the interior of a circle when two chords intersect on the circle

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Can someone please help? its urgent!!!<br><br><br> y=2x+1 for x = -2 , -1 , 0 , 1 , and 2
leonid [27]

Answer:

hello :

y = –2x – 1....(1)

y = 3x–1.....(2)

by (1) and (2) :  3x-1 = -2x-1

5x =0

x =0 ....(the x-coordinates of the solutions)

Step-by-step explanation:

8 0
2 years ago
1. What is a real world example when you would have to find the circumference of something? Explain.
nikitadnepr [17]
A ferris wheel
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4 0
3 years ago
List the domain of the following points from least to greatest: (-2, 4) (-6,8) (3,6) (5,15) (-1, 4)
nasty-shy [4]

Answer:

Domain = {-6, -2, -1, 3, 5}

Step-by-step explanation:

The domain represents the input or x values of a given relation.

In the given set of ordered pairs: (-2, 4) (-6,8) (3,6) (5,15) (-1, 4)

Domain (input/x values) = {-6, -2, -1, 3, 5}

Range (output/y values) = {4, 6, 8, 14}

3 0
3 years ago
Please help?!
Anton [14]

Answer:

Δ PQT ~ Δ QRS  .....{S-S-S test for similarity}...Proof is below.

Step-by-step explanation:

Given:

In Δ PQT

PQ = 30 ft

QT = 28 ft

TP = 20 ft

In Δ QRS

QR = 15 ft

RS = 14 ft

SQ = 10 ft

To Prove:

Δ PQT ~ Δ QRS

Proof:

First we consider  the ratio of the sides

\frac{PQ}{QR}=\frac{30}{15} = \frac{2}{1}            ..............( 1 )

\frac{QT}{RS}=\frac{28}{14} = \frac{2}{1}            ..............( 2 )

\frac{TP}{SQ}=\frac{20}{10} = \frac{2}{1}            ..............( 3 )

So By equation ( 1 ), ( 2 ) and  ( 3 ) we get

\frac{PQ}{QR}=\frac{QT}{RS} = \frac{TP}{SQ}

Now in Δ PQT  and Δ QRS we have

\frac{PQ}{QR}=\frac{QT}{RS} = \frac{TP}{SQ}

Which are corresponding sides of a similar triangle in proportion.

∴ Δ PQT ~ Δ QRS  .....{S-S-S test for similarity}...Proved

8 0
3 years ago
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