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Margarita [4]
3 years ago
5

Why must s-phase occur before mitosis? ...?

Mathematics
1 answer:
stich3 [128]3 years ago
6 0
Here is the answer of the given question above.
S phase happens<span> after G1 </span>phase<span> and this is a crucial period because this is when DNA is copied. G2 phase then happens, which is prior to mitosis. Once this is complete, this then mitosis happens wherein the cells </span><span>divides the contents of the nucleus between two daughter cells. Hope this answer helps.</span>
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9. A circle has an arc of length 56pi that is intercepted by a central angle of 120 degrees. What is the radius of the circle?
SSSSS [86.1K]

Answer:

Part 4) r=84\ units

Part 9) sin(\theta)=-\frac{\sqrt{5}}{3}

Part 10) sin(\theta)=-\frac{9\sqrt{202}}{202}

Step-by-step explanation:

Part 4) A circle has an arc of length 56pi that is intercepted by a central angle of 120 degrees. What is the radius of the circle?

we know that

The circumference of a circle subtends a central angle of 360 degrees

The circumference is equal to

C=2\pi r

using proportion

\frac{2\pi r}{360^o}=\frac{56\pi}{120^o}

simplify

\frac{r}{180^o}=\frac{56}{120^o}

solve for r

r=\frac{56}{120^o}(180^o)

r=84\ units

Part 9) Given cos(∅)=-2/3 and ∅ lies in Quadrant III. Find the exact value of sin(∅) in simplified form

Remember the trigonometric identity

cos^2(\theta)+sin^2(\theta)=1

we have

cos(\theta)=-\frac{2}{3}

substitute the given value

(-\frac{2}{3})^2+sin^2(\theta)=1

\frac{4}{9}+sin^2(\theta)=1

sin^2(\theta)=1-\frac{4}{9}

sin^2(\theta)=\frac{5}{9}

square root both sides

sin(\theta)=\pm\frac{\sqrt{5}}{3}

we know that

If ∅ lies in Quadrant III

then

The value of sin(∅) is negative

sin(\theta)=-\frac{\sqrt{5}}{3}

Part 10) The terminal side of ∅ passes through the point (11,-9). What is the exact value of sin(∅) in simplified form?    

see the attached figure to better understand the problem

In the right triangle ABC of the figure

sin(\theta)=\frac{BC}{AC}

Find the length side AC applying the Pythagorean Theorem

AC^2=AB^2+BC^2

substitute the given values

AC^2=11^2+9^2

AC^2=202

AC=\sqrt{202}\ units

so

sin(\theta)=\frac{9}{\sqrt{202}}

simplify

sin(\theta)=\frac{9\sqrt{202}}{202}

Remember that      

The point (11,-9) lies in Quadrant IV

then      

The value of sin(∅) is negative

therefore

sin(\theta)=-\frac{9\sqrt{202}}{202}

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