X/-3+33=30 solveeeeee

(25)2×100÷23+[24÷(13−5)]

hram777 [196]

Answer:

100*100/23+(24/8)

100*100/23+3

100*4.37+3

43.7+3

46.7

Step-by-step explanation:

7 0

3 years ago

Suppose 15% of the apples picked one afternoon are rotten.

kow [346]

You answer would be Place 20 equally sized pieces of paper in a hat. Of the 20, 3 read "rotten," and the rest read "not rotten". i just took the test

3 0

4 years ago

Read 2 more answers

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

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

5 0

3 years ago

Which could be the name of a plane? please help i will mark brainiest

Scorpion4ik [409]

I think it would be JK but im not really sure

8 0

3 years ago

Find a polynomial f(x) of degree 4 that has the following zeros. 0, -5, 8, -2

Roman55 [17]

Answer:

The polynomial of the function degree '4'

x⁴ - x³ -46 x² -80x =0

Step-by-step explanation:

Explanation:-

Given zeros are 0 ,-5,8,-2

x =0 , x=-5 ,x=8 and x=-2

(x-0) , ( x+5) , (x-8) and (x+2)

The polynomial function with degree '4'

(x-0)(x-(-5)(x-8)(x-(-2) =0

x ( x+5) (x-8) (x+2) =0

(x² +5x)(x-8)(x+2) =0

(x³ -8 x²+5x² -40 x)(x+2) =0

( x³ -3 x² - 40 x) (x +2) =0

x⁴ + 2 x³ - 3 x³- 6 x²-40 x²- 80 x =0

x⁴ - x³ -46 x² -80x =0

Final answer:-

The polynomial of the function degree '4'

x⁴ - x³ -46 x² -80x =0

3 0

3 years ago