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tresset_1 [31]
3 years ago
6

The resultant force, R, is related to two concurrent forces, X and Y, acting at right angles to one another, by the formula R^2=

x^2+y^2 . Rewrite the formula for Y.
Mathematics
2 answers:
vekshin13 years ago
7 0

the answer is y=√r^2-x^2

Korvikt [17]3 years ago
5 0
To find the formula for y in the equation:
r^2=x^2+y^2
First thing we do is we must isolate the y variable
y^2=r^2-x^2
To simplify our answer is we will square root both of the sides of the equation,
y= \sqrt{r^2-x^2}
and this will be the formula for the resultant force for y.
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Transversal t cuts parallel lines a and bas shown in the diagram. Which equation is necessarily true?
irina1246 [14]

Answer:

Option D is correct

Step-by-step explanation:

Using the given diagram, we want to know the equation that is true

Option A is wrong as both are on a straight line and in fact should add up to equal 180 and not be equal to each other

Option B is not correct as both are supplementary and does not equal each other

Option C is not correct, both are corresponding to each other and should not add up to 90

Option D is correct

Both angles are supplementary as they are exterior angles that add up to 180

4 0
3 years ago
The perimeter of a square is 32cm. what is the lenght of each side​
pshichka [43]

Answer:

<h2>s = 8 cm</h2>

Step-by-step explanation:

For a square, P = 4s, where s is the length of one side.

Thus,            32 cm = 4s, and s = 8 cm

8 0
3 years ago
Rewrite with only sin x and cos x. (6 points) cos 2x - sin x
velikii [3]

Answer:

The answer is a. cos2x - sin2x - sin x. Use the rule: cos2x = (cos^2)x - (sin^2)x. So, substitute cos2x in the expression: cos2x - sinx = ((cos^2)x - (sin^2)x) - sinx = (cos^2)x - (sin^2)x - sinx. Therefore, choice a. is correct choice.

4 0
3 years ago
Find the inverse of f(x)=2x-1
solong [7]

{ \qquad\qquad\huge\underline{{\sf Answer}}}

Let's find the inverse of given function ~

\qquad \sf  \dashrightarrow \: f(x) = 2x - 1

\qquad \sf  \dashrightarrow \: 2x = f(x) + 1

\qquad \sf  \dashrightarrow \: x =  \cfrac{f(x) + 1}{2}

[ now, replace f(x) and x with \sf{ {f}^{-1}(x)} ]

\qquad \sf  \dashrightarrow \: f {}^{ - 1} (x) =  \cfrac{x + 1}{2}

That's the required inverse function ~

8 0
1 year ago
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Nataliya [291]
30/100...because it says that it's 30 percent
5 0
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