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2 years ago

G(x)=x^2-5 g(9) help!

Mathematics

1 answer:

emmainna [20.7K]2 years ago

6 0

You’re given g(x) = x^2 - 5

When asked to find g(9), you’re being told that x=9, since the x in the g(x) was replaced by 9.

So that’s telling you to put 9 in for all x’s:
g(9) = (9)^2 - 5

You then just need to clear up the right hand side.

9^2 - 5 = 81 - 5

= 76

So, g(9) = 76

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3w + 7=8 need help with this

LiRa [457]

Hope this helps you out 3w+7=8

The answer is w= 1/3

Step by step:

3w+7=8
Step 1: is subtract is 7 from both sides.
3w+7-7=8-7
3w=1

Step 2: Divide both sides by 3.
3w/3 = 1/3

Your solution is:

w=1/3

6 0

2 years ago

When we use the SPDC model, we always include the following EXCEPT:

Sidana [21]

B.!!!!!! !!!!
that’s the answer

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2 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

2 years ago

Helpp!! Pleaseeeeeeeeeeeeeee

weeeeeb [17]

Answer:

your answer is B i hope this helps :)

Step-by-step explanation:

3 0

2 years ago

Tracy sells water at a fair in 17oz bottles or 11oz cans. If on one shelf she has 11 containers of water that hold a total of 15

blondinia [14]

Answer:

Number of bottles = 6

Number of cans = c

Step-by-step explanation:

Given that:

Container for selling water :

17 oz bottle

11 oz can

Let number of bottles = b

Number of cans = c

b + c = 11 - - - (1)

17b + 11c = 157 - - - (2)

b = 11 - c

Substitute into (2)

17(11 - c) + 11c = 157

187 - 17c + 11c = 157

187 - 6c = 157

-6c = 157 - 187

-6c = -30

c = 30/6

c = 5

From ; b = 11 - c

b = 11 - 5

b = 6

Hence,

Number of bottles = 6

Number of cans = c

3 0

2 years ago