_Brainliest if helped!!

since it varies directly, we form an equation ,
Y =kX where k is a constant
use the points given to find K ,
3 =0.2k , k = 15

So to get final answer,

when x = 1
y = kx, = 1(15)
y= 15

Hence when x=1, < y = 15 >

4 0

3 years ago

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The table below relates the number of rats in a population to time in weeks. Use the table to write a linear equation with w as

salantis [7]

Answer:

C(w) = 6w + 9

Step-by-step explanation:

Anytime 0 is given somewhere, it should be given close scrutiny. In an equation whose general form is

y = mx + b

0 will determine the y intercept immediately.

So when x = 0, y will equal

y = 0*m + 9

So b = 9

y = mx + 9 Now we need to find m

I should start using your variables.

C(w) = m*w + 9

when w = 3 then C(3) = 27

27= 3m +9

27-9 =3m + 9 -9

18 = 3m

18/3 = 3m/3

x = 6

So the complete equation is

C(w) = 6w + 9

7 0

4 years ago

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1440 minutes / 1 day = ? / days

vfiekz [6]

Answer: 1440 minutes is 1 day

Step-by-step explanation:

4 0

3 years ago

Please help me!! 100 points!!

Lyrx [107]

Part A.

We could take, for example:

$\begin{cases}x\geq-3\\y\geq3\end{cases}$

For x ≥ -3 we have vertical line (x-intercept = -3) and shaded region on the right side of the line.
For y ≥ 3 we have horizontal line (y-intercept = 3) and shaded region above that line.

Part B.

Simply, substitute x and y coordinates of D and E <span>to the system of inequalities from part A, and check if its is true, what we get.
</span>
$D=(-2,4)\\\\\begin{cases}x\geq-3\\y\geq3\end{cases}\quad\implies\quad\begin{cases}-2\geq-3\qquad\text{True}\\4\geq3\qquad\text{True}\end{cases}$

so point D is a solution,

$E=(2,4)\\\\\begin{cases}x\geq-3\\y\geq3\end{cases}\quad\implies\quad\begin{cases}2\geq-3\qquad\text{True}\\4\geq3\qquad\text{True}\end{cases}$

and point E is also a solution to <span>our system of inequalities.
</span>
Part C.

There are two ways we could do that. First is the same as in part B. We substitute x and y coordinates of each school and check if inequality is true or false:

$A=(-5,5)\\\\y\ \textless \ 3x-3\quad\implies\quad5\ \textless \ 3\cdot(-5)-3\quad\implies\quad5\ \textless \ -18\quad\text{False}\\\\\\
B=(-4,-2)\\\\y\ \textless \ 3x-3\quad\implies\quad-2\ \textless \ 3\cdot(-4)-3\quad\implies\quad-2\ \textless \ -15\quad\text{False}\\\\\\
C=(2,1)\\\\y\ \textless \ 3x-3\quad\implies\quad1\ \textless \ 3\cdot2-3\quad\implies\quad1\ \textless \ 3\quad\text{True}\\\\\\
D=(-2,4)\\\\y\ \textless \ 3x-3\quad\implies\quad4\ \textless \ 3\cdot(-2)-3\quad\implies\quad4\ \textless \ -9\quad\text{False}\\\\\\
E=(2,4)\\\\y\ \textless \ 3x-3\quad\implies\quad4\ \textless \ 3\cdot2-3\quad\implies\quad4\ \textless \ 3\quad\text{False}\\\\\\$

$F=(3,-4)\\\\y\ \textless \ 3x-3\quad\implies\quad-4\ \textless \ 3\cdot3-3\quad\implies\quad-4\ \textless \ 6\quad\text{True}\\\\\\$

So Timothy can attend schools C and F.

We can also draw a graph of that inequality (pic 2).

8 0

3 years ago

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