All categories

2 years ago

If a pizza with a diameter of 12 inches costs $10.99, based on area, about how much should a 15-inch pizza cost?

Mathematics

1 answer:

Dafna1 [17]2 years ago

5 0

Answer:

Area of a circle=πr^2

12 inch pizza

π(6^2)≈113.04in^2

15 inch pizza

π(7.5^2)≈176.625in^2

Now we turn this into a fraction, and solve the problem:

113.04/10.99=176.625/x

113.04x=1941.10875 <-------Division property of equality

x=17.171875.

The cost of a 15 inch diameter pizza is $17.17

Hope this helped.

You might be interested in

State the slope and y-intercept<br> y=3x+4

WARRIOR [948]

slope intercept form: y=mx+b

y=3x+4

slope is also m

b is the y-intercept

answer:

slope is 3

y-intercept is 4

5 0

2 years ago

Read 2 more answers

SOS HURRY

Burka [1]

First, by using the distance formula for just one side, we can find the length of all sides (a square has 4 equal sides.) Then, we can apply the area of a square formula, which is a^2.

Distance formula:

$\sqrt{(x.2 - x.1)^{2} + (y.2 - y.1)^{2}}$

√((-2 + 5)^2 + (-8 + 4)^2)

√((3)^2 + (4)^2)

√9 + 16

√25

The side lengths of the square are each equal to 5, and by applying the formula for area, we can find the area of the square.

5^2 = 25

3 0

2 years ago

What number must be added to the expression below to complete the square? x2+x

iren2701 [21]

Answer:

answer is 1/4

Step-by-step explanation:

proof is (x-1/2)²= x²-2*1/2*x+1/4= x²-x+1/4

4 0

2 years ago

Is sin theta=5/6, what are the values of cos theta and tan theta?

mina [271]

let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill$

$\bf cos(\theta )=\cfrac{\stackrel{adjacent}{\pm\sqrt{11}}}{\stackrel{hypotenuse}{6}} \\\\\\ tan(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{adjacent}{\pm\sqrt{11}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{tan(\theta )=\pm\cfrac{5}{\sqrt{11}}\cdot \cfrac{\sqrt{11}}{\sqrt{11}}\implies tan(\theta )=\pm\cfrac{5\sqrt{11}}{11}}$

5 0

2 years ago

An object is dropped off a building that us 144 feet tall. After how many seconds does the object hit the ground? (s= 16t^2)

denis23 [38]

Given
s=16t^2
where
s=distance in feet travelled (downwards) since airborne with zero vertical velocity and zero air-resistance
t=time in seconds after release

Here we're given
s=144 feet
=>
s=144=16t^2
=>
t^2=144/16=9
so
t=3
Ans. after 3 seconds, the object hits the ground 144 ft. below.

6 0

3 years ago

Read 2 more answers