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3 years ago

What is the rate of change and initial value for the linear relation that includes the points shown in the table?

Mathematics

1 answer:

melomori [17]3 years ago

5 0

Answer: I need the table to give the answer

Step-by-step explanation: copy and paste the table

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What happens to the function f(x)=x4 when it becomes f′(x)=x4+5?

elena-s [515]

I think it moves 5 units to the right Im not sure tho!!

8 0

2 years ago

Determine the zero of F(x)=x^2+3x-5

IgorLugansk [536]

Answer:

The answer to your question is:

Step-by-step explanation:

x² + 3x - 5

x² + 3x + (3/2)² = 5 + (3/2)²

(x + 3/2)² = 5 + 9/4

(x + 3/2)² = (20 + 9) /4

(x + 3/2)² = 29/4

x + 3/2 = ±√29 / 2

x 1 = -3/2 + √29/2 x2 = -3/2 - √29/2

x1 = 1.19 x2 = -4.19

3 0

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

Sarah is the president of the drama club and wants to publish a newspaper. Expenses are $.90 for printing and mailing each copy,

larisa86 [58]

Let: x = number of copies sold

We can formulate the following equation to solve for x:

1.50x - 0.9x - 600 = 0

x = 1000

Therefore, the club must sell 1000 copies to break even. Among the choices, the correct answer is A.

5 0

3 years ago

(-8)^-2 / 3^-4 simplify the expression. Write your answer using only positiv exponents.

spin [16.1K]

Answer:

3^4 / 8^2 or 81/64

Step-by-step explanation:

(-8)^-2 / 3^-4 =

= 3^4 / (-8)^2

= 3^4 / 8^2

= 81/64

4 0

3 years ago