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3 years ago

PLZZ HELPP IM FAILLING THIS CLASS

Mathematics

1 answer:

arlik [135]3 years ago

4 0

Answer:

Hope this helps!!! :)

Step-by-step explanation:

#1: if there is buy 1 get 1 free, then buy 6 get 3 free. so 2.8 times 3 = $8.4

#2: 4 for $3. break it up to 1 for $0.75. now 0.75 times 6 = $4.5

#3: the second one. $4.5 < $8.4

#4: 45 times 12 = $54

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worty [1.4K]

Answer: -5

Step-by-step explanation: The common difference is the same number we are adding each time to get from one term to the next.

In this sequence, we are adding -5 each time

so our common difference is -5.

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3 years ago

What is the length of side BC of the triangle

noname [10]

The length of side bc is 28

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3 years ago

Hardest math question ever

ohaa [14]

You can name planes by one letter or using three points belonging to it that are <u>not</u> on the same line.

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1 year ago

Considering the expression x/2 + y2. Which statements are true?Mark all that apply.

Svet_ta [14]

Answer:

D. 8/2 +(-10)² = 104

Step-by-step explanation:

To find which statements are true, evaluate the expression for the given variable values. You do this by putting the numbers in place of the respective variables, then doing the arithmetic.

The respective expression values are ...

A 83 ≠ 37 . . . . (4/2) +9² = 2 +81 = 83

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C 52 ≠ 61

D 104 =104 . . . . true statement

I find it less tedious to write a function into a calculator or spreadsheet and let it do the repetitive math. Examples of the calculation are shown above.

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3 years ago

A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.

olga55 [171]

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:

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

7 0

3 years ago