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zhannawk [14.2K]

3 years ago

11

What is the relationship between the ratios? 5/8 and 8/10 Drag and drop to complete the statement. A proportional. B not proport

ional.

2 answers:

Yuliya22 [10]3 years ago

3 0

Answer: A

I hope this helps and have a wonderful day filled with joy!!

finlep [7]3 years ago

3 0

Answer:

it is proportional

Step-by-step explanation:

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Find the area of this figure. Brainliest to first correct answer. Tysm

larisa [96]

Area of Triangle:
A = 1/2 bh = 1/2 (5)(12) = 30 m^2

Area of semicircle:
A = 1/2 π r^2 = 1/2 π 6^2 = 18<span>π m^2

Area of figure = </span>Area of semicircle + Area of Triangle
Area of figure = 18π + 30 m^2

Answer is C.
18π + 30 m^2

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3 years ago

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melamori03 [73]

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3 years ago

Line t intersects line u as shown below <br><br><br> What is the value of x

Rudiy27

Answer:

x = 117 degrees

Step-by-step explanation:

Find the diagram attached

From the diagram, the angle at the top left on line t is equal to theta of the top left on line u and is equal to 117 degrees (corresponding angle)

Also the angle angle on the top left on line u will be equal to x (vertically opposite angle)

Hence the value of x = 117 degrees (vertically opposite to each other on line u)

5 0

3 years ago

What is the lcm of 15 and 10 multiplied by the lcm of 6 and 20

d1i1m1o1n [39]

1800

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LCM of 6 and 20 = 60
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8 0

3 years ago

Which of the following are identities? Check all that apply

Natasha2012 [34]

Answer:

A, C

Step-by-step explanation:

Actually, those questions require us to develop those equations to derive into trigonometrical equations so that we can unveil them or not. Doing it only two alternatives, the other ones will not result in Trigonometrical Identities.

Examining

A) True

$\frac{1-tan^{2}x}{2tanx} =\frac{1}{tan2x} \\ \frac{1-tan^{2}x}{2tanx} =\frac{1}{\frac{2tanx}{1-tan^{2}x}}\\ tan2x=\frac{1-tan^{2}x}{2tanx}$

Double angle $tan2\alpha =\frac{1 -tan^{2}\alpha }{2tan\alpha}$

B) False,

No further development towards a Trig Identity

C) True

Double Angle Sine Formula $sin2\alpha =2sin\alpha *cos\alpha$

$sin(8x)=2sin(4x)cos(4x)\\2sin(4x)cos(4x)=2sin(4x)cos(4x)$

D) False No further development towards a Trig Identity

$[sin(x)-cos(x)]^{2} =1+sin(2x)\\ sin^{2} (x)-2sin(x)cos(x)+cos^{2}x=1+2sinxcosx\\ \\sin^{2} (x)+cos^{2}x=1+4sin(x)cos(x)$

7 0

3 years ago

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