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3 years ago

Let f(x) = twenty four over the quantity of 3 x minus two.Find f(−2). (1 point) Group of answer choices −6 −3 3 6

Mathematics

2 answers:

S_A_V [24]3 years ago

8 0

Answer:

Step-by-step explanation:

<u> 24 </u> =

3x-2

<u> 24 </u>=

3(2)-2

<u> 24 </u>=

6-2

<u> 24 </u>=

Anastasy [175]3 years ago

6 0

Answer:

the answer is 6

Step-by-step explanation:

beccause i took the quiz just now lol

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If you know this pls help ):

Aleksandr-060686 [28]

From the right hand side, we will need to find a way to rewriting 3x²y in terms of cube roots.
We know that 27 is 3³, so if we were to rewrite it in terms of cube roots, we will need to multiply everything by itself two more twice. (ie we can rewrite it as ∛(3x²y)³)

Hence, we can say that it's:
$\sqrt[3]{162x^{c}y^{5}} = \sqrt[3]{(3x^{2}y)^{3}} * \sqrt[3]{6y^{d}}$
$= \sqrt[3]{162x^{6}y^{3+d}}$

Hence, c = 6 and d = 2

4 0

3 years ago

Please Help. I really don’t get this concept, if you could explain it in detail it would be very appreciated

Dafna11 [192]

<h3>Answer: 24/25</h3>

=================================================

Explanation:

Sine is given to be negative, and so is tangent. This only happens in quadrant Q4

Recall that y = sin(theta), so if sin(theta) < 0, then we're below the x axis.

If tan(theta) < 0, then this means cos(theta) > 0

So we have y < 0 and x > 0 which places the angle somewhere in Q4.

--------------------------

Draw a right triangle as shown below in the attached image. We have AC = 25 and BC = 7. Use the pythagorean theorem to find that AB = 24

So this is what your steps may look like

a^2+b^2 = c^2

7^2+b^2 = 25^2

b^2+49 = 625

b^2 = 625-49

b^2 = 576

b = sqrt(576)

b = 24

So because AB = 24, we know that the cosine of the angle is adjacent/hypotenuse = 24/25

---------------------------

As an alternative, you could use the trig identity

sin^2(x) + cos^2(x) = 1

and plug in the given value of sine to solve for cosine. The cosine value result will be positive since we're in Q4.

So,

sin^2(x) + cos^2(x) = 1

(-7/25)^2 + cos^2(x) = 1

(49/625) + cos^2(x) = 1

cos^2(x) = 1 - (49/625)

cos^2(x) = (625/625) - (49/625)

cos^2(x) = (625-49)/625

cos^2(x) = 576/625

cos(x) = sqrt(576/625)

cos(x) = sqrt(576)/sqrt(625)

cos(x) = 24/25

This is effectively a rephrasing of the previous section since the pythagorean trig identity is more or less the pythagorean theorem (just in a trig form)

6 0

3 years ago

Expression for all solutions to the equation cos(theta) = 1 + 3cos(theta)

OleMash [197]

$\cos\theta=1+3\cos\theta\implies 0=1+2\cos\theta\implies \cos\theta=-\dfrac12$

On the unit cirlce, the cosine of an angle is negative whenever the angle falls in the interval $\dfrac\pi2$ .

If you recall your special triangles, this cosine occurs for angles of $x=\dfrac{2\pi}3$ and $x=\dfrac{4\pi}3$ .

More generally, this occurs for $x=\dfrac{2\pi}3+2n\pi$ and $x=\dfrac{4\pi}3+2n\pi$ , where $n$ is an integer.