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3 years ago

Solve s = rθ for θ. A) θ = rs B) θ = r/s C) θ = s/r D) θ = 1/rs

Mathematics

2 answers:

meriva3 years ago

5 0

Dividing both sides by r

=s/r

Schach [20]3 years ago

5 0

A) i believe it is a hope this helps

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F(x) = 3x - 3 <br> g(x) = 4x+5 <br> find (f•g) (3) <br><br> only numerical answers work

Andreas93 [3]

Answer:

48 bcoz u do 4(3)+5 do f(g) than for f(17) u do 3(17)-3

6 0

3 years ago

Suppose the graph of a cubic polynomial function has the same zeroes and passes through the coordinate (0, -5). Write the equati

amid [387]

In this we know all three zeros and one point from which the graph pass.
So we will let specific cubic polynomial function of the form
$f(x) = a(x - x_1)(x-x_2)(x-x_3)$

As we know zeros are that point where we will get value of function equal to zero. So it is basically in form $(x_n , 0)$

SO in given question zeros are (2 , 0) , (3, 0) and (5,0)
So we can say $x_1 = 2 , x_2 = 3 , x_3 = 5$

So required equation is
$f(x) = a (x-2)(x-3)(x-5)$
              $= a[(x^2 - 2x - 3x + 6)(x-5)]$
              $= a[(x^2 - 5x+6)(x-5)]$
              $= a(x^3 - 5x^2 + 6x- 5x^2 + 25x - 30)$
              $= a(x^3 - 10x^2+31x-30)$
Now we have one point (0 , -5) from which graph passes.
So we say at x = 0 , f(x) = -5
$-5= a (0-0+0 - 30)$
$-5 = -30a$
$a = \frac{-5}{-30} = \frac{1}{6}$
So required equation of cubic polynomial is
$f(x) = \frac{1}{6}(x^3 -10x^2+31x-30)$

For finding y - intercept we simply plugin x = 0 in given equation.
As we know at x = 0 , value of function is -5.
So y - intercept is -5.

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3 years ago

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Please help me I will mark as a brainliest

MrRissso [65]

Answer:

grv srt66as 8f bolly fun. bolly. bolly fun tv

Step-by-step explanation:

vr tvtybin I'll wind Erin posh key tv I slap tv

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2 years ago

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alani wants to buy a $360 bicycle. she is considering two payment options the image shows option A which consists of making an i

Nataliya [291]

for A)

if she has $ for the downpayment of $70

for B)

360-70+290

290/ 5=58

amount paid = 290-58(#month)

a=290-58m

ANSWER

A= -58m+290

8 0

3 years ago

How can the Pythagorean Identity be used to find sin θ, cos θ, or tan θ and the quadrant location of the angle?

Archy [21]

First lets see the pythagorean identities

$sin^2 \Theta + cos^2 \Theta =1$

So if we have to solve for sin theta , first we move cos theta to left side and then take square root to both sides, that is

$sin^2 \Theta = 1-cos^2 \Theta => sin \theta = \pm \sqrt{1-cos^2 \Theta}$

Now we need to check the sign of sin theta

First we have to remember the sign of sin, cos , tan in the quadrants. In first quadrant , all are positive. In second quadrant, only sin and cosine are positive. In third quadrant , only tan and cot are positive and in the last quadrant , only cos and sec are positive.

So if theta is in second quadrant, then we have to positive sign but if theta is in third or fourth quadrant, then we have to use negative sign .

5 0

3 years ago