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3 years ago

Find the circumference and the area of a circle with radius 5 yd

Mathematics

1 answer:

slavikrds [6]3 years ago

6 0

Answer:

C = 10(pi) yards

A = 25(pi) yards

Step-by-step explanation:

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Write the equation of a line in standard form that passes through (-2, 5) and is perpendicular to

sdas [7]

Answer:

Step-by-step explanation:

x - 2y = 8

Write this in y = mx +b form

- 2y = -x + 8

Divide the equation by (-2)

$\dfrac{-2y}{-2}=\dfrac{-x}{-2}+\dfrac{8}{-2}\\\\\\y =\dfrac{1}{2}x-4$

Slope = 1/2

The slope of the perpendicular line = -1/m

$=\dfrac{-1}{\dfrac{1}{2}}=1*\dfrac{-2}{1} = -2$

m = -2 , ( -2 , 5)

Equation: y = mx +b

Plug in m = -2 ,x = -2 and y =5

5 = (-2)*(-2) + b

5 = 4 + b

5 - 4 = b

b = 1

Equation of the line:

y = mx +b

y = -2x + 1

7 0

3 years ago

Pls help my teacher is so rude

storchak [24]

The third one hope this helps

3 0

3 years ago

I’m sorry if I’m asking for too much but this whole page is very confusing And I don’t remember learning this in fifth grade hel

nekit [7.7K]

Answer:

1.) 71 > 38 ║d.) >

2.) 21 < 62 ║g.) <

3.) 58 = 58 ║f.) =

4.) 14 ≤ 6 ║a.) False

5.) 13 ≤ 26 ║h.) True

6.) 25 < y ║c.) 16

7.) N = 6 ║e.) 6

8.) P ≥ 16 ║b.) 50

8 0

3 years ago

What is the slope? Pls help thanks I think it’s 0 but not sure

kondaur [170]

Yeah I think your right but I’m not 100 percent sure either

7 0

3 years ago

Read 2 more answers

Evaluate cosA/2 given cosA=-1/3 and tanA >0

mote1985 [20]

Answer:

$\bold{cos\dfrac{A}{2} = -\dfrac{1}{\sqrt3}}$

Step-by-step explanation:

Given that:

$cosA=-\dfrac{1}3$

and

$tanA > 0$

To find:

$cos\dfrac{A}{2} = ?$

Solution:

First of all,we have cos value as negative and tan value as positive.

It is possible in the 3rd quadrant only.

$\dfrac{A}{2}$ will lie in the 2nd quadrant so $cos\dfrac{A}{2}$ will be negative again.

Because Cosine is positive in 1st and 4th quadrant.

Formula:

$cos2\theta =2cos^2(\theta) - 1$

Here $\theta = \frac{A}{2}$

$cosA =2cos^2(\dfrac{A}{2}) - 1\\\Rightarrow 2cos^2(\dfrac{A}{2}) =cosA+1\\\Rightarrow 2cos^2(\dfrac{A}{2}) =-\dfrac{1}3+1\\\Rightarrow 2cos^2(\dfrac{A}{2}) =\dfrac{2}3\\\Rightarrow cos(\dfrac{A}{2}) = \pm \dfrac{1}{\sqrt3}$

But as we have discussed, $cos\dfrac{A}{2}$ will be negative.

So, answer is:

$\bold{cos\dfrac{A}{2} = -\dfrac{1}{\sqrt3}}$

7 0

3 years ago