One size of pizza served at Joe’s Pizza Parlor is 10in. in diameter. What is the area of this particular (circular) size pizza? (Pie=3.14)

Answers

Answer 1
Answer:

Answer:

78.5 in.²

Explanation:

Formula For Area Of Circle: Area = πr²

π: 3.14

r (Half Of Diameter): 5

r²: 25

3.14 · 25 = 78.5


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A stock of food is enough to fees 50 persons for 14 days. How many days will the foos last if 20 persons will be added?

Answers

Answer:

solution

Step-by-step explanation:8 days

Sophia earned $240 at her brother's store. If she worked for 8 days and earned the same amount of money each day, how much did she earn per day?

Answers

Answer:

$30

Step-by-step explanation:

$240 divided by 8

She earns $30 each day

8x=240

Where 8 is the amount of days, 240 is the total pay, and x represents the pay per day.

8x=240
—- ——
8 8

In this step, you are dividing both sides by 8 to isolate the x

x=30

Therefore, she was paid $30 each day. You can check this by inserting back into your equation.

8(30)=240
240=240

Hope this helps!

Katelyn had 2 dogs and 3 cats. Which of the following shows the ratio written correctly for the number of dogs to cats?A) 3:2
B) 3-2
C) 2:3
D) 3 x 2

Answers

Answer:

2:3

Step-by-step explanation:

Ratio is indicated by :

Its dogs to cats, right so ratio is number of dogs:number of cats which is 2:3

Hope this helps plz mark brainliest if correct :D

Answer:

2:3

Step-by-step explanation:

2 is first and 3 is the second number so you put them in order

Scarlett is playing a video game. She spends 900 minerals to create 18 workers. Each worker costs the same number of minerals.

Answers

Answer: m=50w

Step-by-step explanation:

When you divide 900 by 18, you get 50.

Since we are trying to find the amount of minerals per worker, m is the first variable. w is the second because that is the number of workers you can get with the minerals Scarlett has.

So you would do 900÷18 to get the amount each worker is worth, which would be 50 minerals.

Pleeease open the image and hellllp me

Answers

1. Rewrite the expression in terms of logarithms:

y=x^x=e^(\ln x^x)=e^(x\ln x)

Then differentiate with the chain rule (I'll use prime notation to save space; that is, the derivative of y is denoted y' )

y'=e^(x\ln x)(x\ln x)'=x^x(x\ln x)'

y'=x^x(x'\ln x+x(\ln x)')

y'=x^x\left(\ln x+\frac xx\right)

y'=x^x(\ln x+1)

2. Chain rule:

y=\ln(\csc(3x))

y'=\frac1{\csc(3x)}(\csc(3x))'

y'=\sin(3x)\left(-\cot^2(3x)(3x)'\right)

y'=-3\sin(3x)\cot^2(3x)

Since \cot x=(\cos x)/(\sin x), we can cancel one factor of sine:

y'=-3(\cos^2(3x))/(\sin(3x))=-3\cos(3x)\cot(3x)

3. Chain rule:

y=e^{e^(\sin x)}

y'=e^{e^(\sin x)}\left(e^(\sin x)\right)'

y'=e^{e^(\sin x)}e^(\sin x)(\sin x)'

y'=e^{e^(\sin x)+\sin x}\cos x

4. If you're like me and don't remember the rule for differentiating logarithms of bases not equal to e, you can use the change-of-base formula first:

\log_2x=(\ln x)/(\ln2)

Then

(\log_2x)'=\left((\ln x)/(\ln 2)\right)'=\frac1{\ln 2}

So we have

y=\cos^2(\log_2x)

y'=2\cos(\log_2x)\left(\cos(\log_2x)\right)'

y'=2\cos(\log_2x)(-\sin(\log_2x))(\log_2x)'

y'=-\frac2{\ln2}\cos(\log_2x)\sin(\log_2x)

and we can use the double angle identity and logarithm properties to condense this result:

y'=-\frac1{\ln2}\sin(2\log_2x)=-\frac1{\ln2}\sin(\log_2x^2)

5. Differentiate both sides:

\left(x^2-y^2+\sin x\,e^y+\ln y\,x\right)'=0'

2x-2yy'+\cos x\,e^y+\sin x\,e^yy'+\frac{xy'}y+\ln y=0

-\left(2y-\sin x\,e^y-\frac xy\right)y'=-\left(2x+\cos x\,e^y+\ln y\right)

y'=(2x+\cos x\,e^y\ln y)/(2y-\sin x\,e^y-\frac xy)

y'=(2xy+\cos x\,ye^y\ln y)/(2y^2-\sin x\,ye^y-x)

6. Same as with (5):

\left(\sin(x^2+\tan y)+e^(x^3\sec y)+2x-y+2\right)'=0'

\cos(x^2+\tan y)(x^2+\tan y)'+e^(x^3\sec y)(x^3\sec y)'+2-y'=0

\cos(x^2+\tan y)(2x+\sec^2y y')+e^(x^3\sec y)(3x^2\sec y+x^3\sec y\tan y\,y')+2-y'=0

\cos(x^2+\tan y)(2x+\sec^2y y')+e^(x^3\sec y)(3x^2\sec y+x^3\sec y\tan y\,y')+2-y'=0

\left(\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^(x^3\sec y)-1\right)y'=-\left(2x\cos(x^2+\tan y)+3x^2\sec y\,e^(x^3\sec y)+2\right)

y'=-(2x\cos(x^2+\tan y)+3x^2\sec y\,e^(x^3\sec y)+2)/(\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^(x^3\sec y)-1)

7. Looks like

y=x^2-e^(2x)

Compute the second derivative:

y'=2x-2e^(2x)

y''=2-4e^(2x)

Set this equal to 0 and solve for x :

2-4e^(2x)=0

4e^(2x)=2

e^(2x)=\frac12

2x=\ln\frac12=-\ln2

x=-\frac{\ln2}2

You can run 10 miles in 80 minutes. How far can you run in 2.5 hours?

Answers

Answer:

18.75 miles

Step-by-step explanation:

1 hour = 60 minutes

2.5 Hours = 150 minutes

10 miles _____> 80 minutes

x_______> 150 minutes

x= ( 150 * 10) / 80

= 1500/80

x= 18.75 miles

easy

Answer:

Step-by-step explanation:

10 miles- 80 minutes

            ?- 150 minutes

1500+80x

X= 10.75 miles