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RUDIKE [14]
2 years ago
10

I will mark brainly, help me with this question please.​

Mathematics
1 answer:
Anna [14]2 years ago
3 0

Step-by-step explanation:

Since x varies directly as \sqrt{y} we can write the relation as

x = k\sqrt{y}

where k is the constant of proportionality.

a) To solve for the k, we substitute the given values:

5 = k\sqrt{13} = k(13)

\Rightarrow k = \dfrac{5}{13}

b) The equation relation x and y can be written as

x = \dfrac{5}{13}\sqrt{y}

c) When y = 9,

x = \dfrac{5}{13}\sqrt{9} = \dfrac{5}{13}(3) = \dfrac{15}{13}

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Solve the equation and give me how you solved it and the answer
Viktor [21]
Multiplying both sides by 3, we get (3/9)*(2m-16)=(2m+4). Next, we multiply both sides by 9 and expand to get 6x-48=18m+36. Subtracting 6x and 36 from both sides, we get 12x=-48-36=-84. After that, we divide both sides by 12 to get m=-7
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)

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ivolga24 [154]
<span>lIt occurs in the left parietal lobe. 

Happy studying ^-^</span>
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