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3 years ago

What is the scale factor?

Mathematics

2 answers:

liraira [26]3 years ago

4 0

Answer:

3

Step-by-step explanation:

As you can see we can compare one side of the larger triangle to the other one by seeing the length of the side

One side measures 27 while another one measures 9 (from the other triangle) to get to 9 starting by 27, we divide by 3…

Alexandra [31]3 years ago

3 0

Answer: B

Step-by-step explanation:

You can see that the length of the sides from abc are factored by 1/3 to get the length of xyz. Therefore, the scale factor is 1/3.

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sergey [27]

The answer is 7 i did it on a calculator lol

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3 years ago

HELP!!!! Position and label the triangle on the coordinate plane

Darya [45]

Answer:

Option (2).

Step-by-step explanation:

It is given in the question,

ΔLMN is a right triangle with base LM = 3a units

Hypotenuse MN = 5a

By applying Pythagoras theorem in ΔLMN,

MN² = LM² + NM²

(5a)² = (3a)² + MN²

25a² - 9a² = MN²

MN = √16a²

MN = 4a

Therefore, vertices of the triangle will be L(0, 0), M(3a, 0) and N(0, 4a).

Option (2) will be the answer.

4 0

3 years ago

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lora16 [44]

Answer:

1st, 3rd and 4th

Step-by-step explanation:

Only 2nd is false

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4 years ago

8x+4y=-2<br> -12x+5y=-8<br> What is the answer

tester [92]

Answer:

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3 years ago

Given: KL ║ NM , LM = 45, m∠M = 50° KN ⊥ NM , NL ⊥ LM Find: KN and KL

Mice21 [21]

Answer:

$KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08\\ \\KN=45\sin 50^{\circ}\approx 34.47$

Step-by-step explanation:

Given:

KL ║ NM ,

LM = 45

m∠M = 50°

KN ⊥ NM

NL ⊥ LM

Find: KN and KL

1. Consider triangle NLM. This is a right triangle, because NL ⊥ LM. In this triangle,

LM = 45

m∠M = 50°

So,

$\tan \angle M=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{NL}{LM}=\dfrac{NL}{45}\\ \\NL=45\tan 50^{\circ}$

Also

$m\angle LNM=90^{\circ}-50^{\circ}=40^{\circ}$ (angles LNM and M are complementary).

2. Consider triangle NKL. This is a right triangle, because KN ⊥ NM . In this triangle,

$NL=45\tan 50^{\circ}$

$m\angle KLN=m\angle LNM=40^{\circ}$ (alternate interior angles)

$m\angle KNL=90^{\circ}-40^{\circ}=50^{\circ}$ (angles KNL and KLN are complementary).

So,

$\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08$

and

$\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47$

3 0

3 years ago

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