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melamori03 [73]
3 years ago
15

What is 3(1+6r)=14-12

Mathematics
1 answer:
olasank [31]3 years ago
7 0

Answer:

<em>r = 0.055555.....</em>

Step-by-step explanation:

3(1+6r)=14-12        Write out formula.

3+18r=2                Distributive Property Form.

(3-3)+18r=2-3        Subtract 3 to both sides.

(18/18)r=-1/18         Divide both sides by 18

r=0.055555..... or -1/18.                    

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The sum of two consecutive odd integers is 20 less than four times the smaller of the two integers. Which the following equation
Rudiy27

Answer:

consecutive = x+x+2

x+x+2 = 4x-20

2x+2=4x-20

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22=2x

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Hope this helps plz hit the crown :D

8 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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