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3 years ago

Help me all please!!

Mathematics

2 answers:

Licemer1 [7]3 years ago

8 0

Well every pizza cost 9.50. So to see the cost for 2 pizzas you can do 2 * 9.50 and for 3 pizzas would be 3 * 9.50 and for 4 is 4 * 9.50 and this goes on forever. notice how the number of pizzas is the only thing that changes. So so we could substitute p for the number of pizzas and say
p pizzas cost p * 9.50 or 9.50p

To have it delivered there is a flat rate of 4.50. this can be added onto the price of p pizzas and we get
cost is 9.50p + 4.50 or
c = 9.50p + 4.50

2. h + 7 =15. in this equation we want to get the h by itself to its value. well to get the h by itself I have to get rid of the + 7. To to make + 7 = 0 I would have to put a - 7. But to put a -7 on one side of an equation i have to put it on other side like this:
h + 7 - 7 = 15 - 7
h + 0 = 8
h = 8

3. same as above but addition
m - 10 + 10 = 24 + 10
m = 34

4. same but division
6 * t / 6 = 96 / 6
t = 16

5. same but multiplication
p/7 *7 =3 * 7
p = 21

Harlamova29_29 [7]3 years ago

7 0

No matter how many pizzas he orders (P) he will be charged 4.50. then each pizza is worth 9.50 so you multiply that by how many he orders (5)
c=4.5+9.5p
c=4.5+9.5(5)
c=$52

for these problems you do the opposite of what is shown
1. h+7=15, subtract 7 from each side. h=8
2. m-10=24, add 10 to each side m=34
3.6t=96, divide both sides by 6 t=16
4. multiply both sides by 7, p=21.

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We know that:

$(x-a)^2+(y-b)^2=r^2$

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}$

and all we have to do is solve it for a, b and r.

There will be:

$\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\
\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\
$

From equations (II) and (III) we have:

$\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}$

and from (I) and (II):

$\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}$

Now we can easly calculate a and b:

$\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}$

Finally we calculate $r^2$ :

$a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}$

And the equation of the circle is:

$(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}$

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