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kvasek [131]
3 years ago
11

Draw scalene triangle ABC and extend side BC to the right of vertex C.

Mathematics
1 answer:
Doss [256]3 years ago
3 0
Mark Brainliest please


See the image for the drawing
Let’s say the side BC is extended till D

1) measure of angle ACD = 105° ( angle ACD is exterior angle)
45 + 60 = 105
2) I found the answer by adding the values of Angle A and Angle B or by using the exterior angle property.
3) I observed that exterior angle property can be used here. Exterior angle property is that an exterior angle of a triangle is equal to the sum of the opposite interior angles.

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What is 0.1 kiloliters to liters?​
konstantin123 [22]

Answer: 100

Multiply the volume value by 1000

.1×1000=100

Move the decimal to the right since your multiplying.

Since multiplying 1000 move it 3 times.

After you move it 3 times put 2 zeros behind 1.

You should get 100.

5 0
4 years ago
Plz help <br> People with a strong mind won't ignore
Genrish500 [490]

Answer:

28. (H) 24 ft

30. (F) \frac{5}{52}

Step-by-step explanation:

28. A right triangle can be formed, where the length of the cable represents the hypotenuse, and the ground represents the base of the triangle. From the pythagorean theorem,

a^2 + b^2 = c^2

where a and b are legs, and c is the hypotenuse of a right triangle.

So

18^2 + b^2 = 30^2\\b = \sqrt{30^2 - 18^2}\\b = \sqrt{576}\\b = 24

30. The probability of drawing a red 1 is \frac{2}{52} (a red star 1 or a red heart 1), the probability of drawing a black 3 is \frac{2}{52} (a black triangle 3 or a black circle 3), and the probability of drawing a six of hearts is \frac{1}{52} (since there is only one six of hearts in the entire deck). Since each of the cards is mutually exclusive from one another, the probability of drawing any of them is simply all of their probabilities added together. That is,

\frac{2}{52} + \frac{2}{52} + \frac{1}{52} = \frac{5}{52}

6 0
4 years ago
Given that SQ¯¯¯¯¯ bisects ∠PSR and ∠SPQ≅∠SRQ, which of the following proves that PS¯¯¯¯¯≅SR¯¯¯¯¯?
yanalaym [24]

Step-by-step explanation:

It is given that ∠SPQ≅∠SRQ

.

The figure shows the same triangles P Q S and R Q S as in the beginning of the task. Angles S P Q and S R Q are highlighted in red.

It is also given that SQ⎯⎯⎯⎯⎯

bisects ∠PSR

.

By the definition of angle bisector, ∠PSQ≅∠QSR

.

The figure shows the same triangles P Q S and R Q S as in the beginning of the task. Angles P S Q and R S Q are congruent and highlighted in red.

△PQS

and △RQS share a common side SQ⎯⎯⎯⎯⎯, and SQ⎯⎯⎯⎯⎯≅SQ⎯⎯⎯⎯⎯

by the Reflexive Property of Congruence.

The figure shows the same triangles P Q S and R Q S as in the beginning of the task. Segment S Q is highlighted in red.

Two angles, ∠SPQ

and ∠PSQ, and a nonincluded side, SQ⎯⎯⎯⎯⎯, of △PQS are congruent to two angles, ∠SRQ and ∠QSR, and a nonincluded side, SQ⎯⎯⎯⎯⎯, of △RQS

.

The figure shows the same triangles P Q S and R Q S as in the beginning of the task. Angles P S Q and R S Q are highlighted in red. Angles S P Q and S R Q are highlighted in red. Side S Q is highlighted in blue.

So, △PQS≅△RQS

by the Angle-Angle-Side (AAS) Congruence Theorem.

PS⎯⎯⎯⎯⎯

and SR⎯⎯⎯⎯⎯ are corresponding sides of congruent triangles, △PQS and △RQS. So, PS⎯⎯⎯⎯⎯≅SR⎯⎯⎯⎯⎯

by CPCTC.

The figure shows the same triangles P Q S and R Q S as in the beginning of the task. Sides S R and S P are congruent and highlighted in red.

Translate these six statements and reasons into a 2

-column proof,

1. ∠SPQ≅∠SRQ

(Given)

2. SQ⎯⎯⎯⎯⎯

bisects ∠PSR

. (Given)

3. ∠PSQ≅∠QSR

(Def. of ∠

bisect)

4. SQ⎯⎯⎯⎯⎯≅SQ⎯⎯⎯⎯⎯

(Reflex. Prop. of ≅

)

5. △PQS≅△RQS

(AAS Steps 1, 3, 4)

6. PS⎯⎯⎯⎯⎯≅SR⎯⎯⎯⎯⎯

(CPCTC)

There you go

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3 years ago
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6 0
3 years ago
SOMEONE PLEASE JUST ANSWER THIS FOR BRAINLIEST!!!
Inga [223]

Remember to be careful with signs when adding and subtracting. -g^4-10g^2h^2-3h^4

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