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aleksandrvk [35]
3 years ago
5

Using what you know about similar figure, solve for X

Mathematics
1 answer:
ale4655 [162]3 years ago
3 0
Since the scale factor is 3:1 you can multiply the length of the side of the triangle to the right by 3 to make the ratio 3:3

the equation would be:
2x+5 = 45

subtract 5 from both sides
2x = 40

divide both sides by 2

x = 20

hope this helps

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Interquartile range of 6 26 27 28 29 30 32 36 38 40
geniusboy [140]
The first quartile is 27 and the third is 36 you would subtract 36 - 27 = 9.
7 0
4 years ago
I need help<br> 4x^2-8x-23=0
Strike441 [17]

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Answer: Check the screenshot below!

Explanation:

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8 0
3 years ago
PLEASE ANSWER!!! NEED HELP WITH EXPLANATION!!!
tiny-mole [99]

Answer:

27.37 and 9.18

Step-by-step explanation:

(4)

Using the cosine ratio in the right triangle

cos64° = \frac{adjacent}{hypotenuse} = \frac{12}{x} ( multiply both sides by x )

x × cos64° = 12 ( divide both sides by cos64° )

x = \frac{12}{cos64} ≈ 27.37 ( to the nearest hundredth )

(5)

Using the sine ratio in the right triangle

sin35° = \frac{opposite}{hypotenuse} = \frac{x}{16} ( multiply both sides by 16 )

16 × sin35° = x , then

x ≈ 9.18 ( to the nearest hundredth )

5 0
3 years ago
Ms. Ramos thumb measures 4 centimeters. Express this length in meters
Mademuasel [1]
This length is 1/25 (0.04) meters.
7 0
4 years ago
Read 2 more answers
PLEASE HELP Given: △KLM LM=12, m∠K=60°, m∠M=45° Find: Perimeter of △KLM.
worty [1.4K]

We have to find the perimeter of the triangle KLM.

We have been given that the length of the side LM=12, m\angleK=60^\circ, and m\angle M= 45^\circ

Refer the attached image.

In a triangle sum of three angles should be 180^\circ.

So,

m\angle K+m\angle L+m\angle M=180^\circ

Plugging the values of angle K and angle M, we get:

60^\circ+m\angle L+45^\circ=180^\circ

So,

m\angle L=180^\circ-105^\circ=75^\circ

Now, that we have the measure of angle L, we will apply sine rule to find the length of the sides KL and KM.

Using the sine law for the triangle KLM, we get:

\frac{sin K}{LM}=\frac{sin L}{KM}=\frac{sin M}{KL}

Refer the image. Plugging the value of the sides of the triangle KLM and the angles of the triangle KLM, we get:

\frac {sin 60^\circ}{12}=\frac{sin 75^\circ}{y}=\frac{sin 45^\circ}{x}

Now using,

\frac {sin 60^\circ}{12}=\frac{sin 75^\circ}{y}

We get the value of 'y'

y=\frac{sin 75^\circ}{sin 60^\circ} \times 12=\frac{0.9659}{0.866} \times 12=13.38

So the length of the side KM is 13.38 units.

Now using,

\frac {sin 60^\circ}{12}=\frac{sin 45^\circ}{x}

We get the value of 'x'

x=\frac{sin 45^\circ}{sin 60^\circ} \times 12=\frac{0.707}{0.866} \times 12=9.79

So the length of the side KL is 9.79 units.

Now, to find the perimeter of the triangle KLM we need to sum up the length of the sides of the triangle KLM.

The perimeter of the triangle KLM = KL+ LM + KM = 9.79 + 12 + 13.38 = 35.17 units

6 0
3 years ago
Read 2 more answers
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