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3 years ago

Angle VSU and TSR are what kind of angles?

Mathematics

1 answer:

Anit [1.1K]3 years ago

6 0

In this question, we have two angles given, which are

$\angle VSU \ and \ \angle TSR$

And we have to find the relationship between those two angles .

Let's first check what are Vertically Opposite Angles

They are the angles opposite to each other when two lines crossing each other .

And the given angles are formed by the two lines crossing each other and these angles are opposite to each other. So these angles are Vertically Opposite Angles .

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Find the exact value of sin (x-y) if sinx=4/9 and siny=1/4

Svetach [21]

Answer:

sin(x - y) = 0.21

Step-by-step explanation:

we have the sin values which we need to get cos values

sin (A-B) = sin A cos B - sin B cos A

sin² A + cos² A = 1

sin x = 4/9

cos² x = 1 - sin² x = 1 - 16/81 = 65/81

cos² x = 65/81

cos x = √65/9

sin y = 1/4

cos² y = 1 - sin² y = 1 - 1/16 = 15/16

cos² y = 15/16

cos y = √15/4

sin(x − y) = sin x cos y - sin y cos x

sin(x - y) = 4/9 √15/4 - 1/4 √65/9

sin(x - y) = (4√15-√65)/36

sin(x - y) = 0.21

socratic Narad T

8 0

2 years ago

Suppose you deposited $900 into an account that earns 4% simple annual interest. Find the balance of the account after 6 months

Jet001 [13]

Balance after 6 months would be 900*(900+2%). You just need to evaluate it then you will get the answer. btw 2% comes from 4% because it's half of the year.

3 0

3 years ago

Which expression has a value of 24 when x=4 and y=2?

wolverine [178]

C) will have value of 24

4 0

4 years ago

Read 2 more answers

Solve the equation :<br>-3 • ( 2 - x ) + 4 = 2 • ( 1 - 2x) + 3<br>thanks :)

kakasveta [241]

Isolate the variable by dividing each side by factors that don't contain the variable.

x = 1

-6+3x+4=2-4x+3

-8+7x+1=0

7x=7

x=1

4 0

3 years ago

What is the justification for the first step in proving the formula for factoring the sum of cubes?

Anna11 [10]

Explanation:

The formula isnt correctly written, it should state:

$a^3+b^3 = (a+b)(a^2-ab+b^2)$

You have to start from $(a+b)(a^2-ab+b^2)$ and end in a³+b³. On your first step, you need to use the distributive property.

$(a+b)(a^2-ab+b^2) = a*(a^2-ab+b^2) + b*(a^2-ab+b^2)$

This is equal to

$a*a^2-a*(ab) + a*b^2 + b*a^2-b*(ab) + b*b^2 = a^3 - a^2b + ab^2 +ba^2 -b^2a +b^3$

Note that the second term, -a²b, is cancelled by the fourth term, ba², and the third term, ab², is cancelled by the fifht term, -b²a. Therefore, the final result is a³+b³, as we wanted to.

5 0

4 years ago