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svetoff [14.1K]
2 years ago
7

Graph: y = (x – 3)2 + 3

Mathematics
1 answer:
dangina [55]2 years ago
5 0

Answer:

Use, y = 3x-3, it might be easier to graph

Step-by-step explanation:

You can convert this equation into slope-intercept for to gramph.

y = (x – 3)2 + 3

y = 3x-6+3

y = 3x-3

You might be interested in
What is the product of -7/10 and 1/2?
Airida [17]
-7/20
Hope I helped :)
6 0
3 years ago
if a rectangle has am area of 2x²+11x+15, what would be an expression for the length, and for the width?
lozanna [386]

Answer:

Length = 2x + 5

Width = x + 3

Step-by-step explanation:

Area of rectangle = length × width

Expression for area of the rectangle = 2x² + 11x + 15

Factorising the quadratic expression

2x² + 11x + 15 = 2x² + 6x + 5x + 15 = (2x² + 6x) + (5x + 15) = 2x(x + 3) +5(x + 3) = (2x + 5)(x + 3)

Length = 2x + 5

Width = x + 3

3 0
2 years ago
A tank with a capacity of 1000 L is full of a mixture of water and chlorine with a concentration of 0.02 g of chlorine per liter
faltersainse [42]

At the start, the tank contains

(0.02 g/L) * (1000 L) = 20 g

of chlorine. Let <em>c</em> (<em>t</em> ) denote the amount of chlorine (in grams) in the tank at time <em>t </em>.

Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of

(<em>c</em> (<em>t</em> )/(1000 + (10 - 25)<em>t</em> ) g/L) * (25 L/s) = 5<em>c</em> (<em>t</em> ) /(200 - 3<em>t</em> ) g/s

In case it's unclear why this is the case:

The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after <em>t</em> s there will be (1000 + 10<em>t</em> ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time <em>t</em> is (1000 - 15<em>t </em>) L. Then the concentration of chlorine per unit volume is <em>c</em> (<em>t</em> ) divided by this volume.

So the amount of chlorine in the tank changes according to

\dfrac{\mathrm dc(t)}{\mathrm dt}=-\dfrac{5c(t)}{200-3t}

which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5<em>t</em> )^(5/3), then integrate both sides to solve for <em>c</em> (<em>t</em> ):

\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{200-3t}=0

\dfrac1{(200-3t)^{5/3}}\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{(200-3t)^{8/3}}=0

\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0

\dfrac{c(t)}{(200-3t)^{5/3}}=C

c(t)=C(200-3t)^{5/3}

There are 20 g of chlorine at the start, so <em>c</em> (0) = 20. Use this to solve for <em>C</em> :

20=C(200)^{5/3}\implies C=\dfrac1{200\cdot5^{1/3}}

\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}

7 0
3 years ago
Endpoint given the Midpoint: * The midpoint of RP is M(2, 4). If one of the end points is R(-1,7), find the coordinates of the o
marin [14]

Answer:

The coordinates of other point are: (5,1)

Step-by-step explanation:

Given coordinates are:

M(x,y) = (2,4)

R(x_1,x_2) = (-1,7)

We have to find the coordinates of other point (x2,y2)

The formula for mid-point is given by:

M(x,y) = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})

Putting the values we get

(2,4) = (\frac{-1+x_2}{2}, \frac{7+y_2}{2})

Putting respective elements equal

\frac{-1+x_2}{2} = 2\\-1+x_2 = 2*2\\-1+x_2 = 4\\x_2 = 4+1\\x_2 = 5\\And\\\frac{7+y_2}{2} = 4\\7+y_2 = 4*2\\7+y_2 = 8\\y_2 = 8-7\\y_2 = 1

Hence,

The coordinates of other point are: (5,1)

8 0
2 years ago
I am between 24 and 34 u say my name when u count by twos from 0 u say my name when u count by 5 from 0 what number am i
katen-ka-za [31]
Hi there 
let me help you out 

The answer is 30 
To find the answer you have to find the multiples of 2 and 5 
2,4,6,8,10,12, 14,16,18,20,22,24,26,28,30,32,34,
5,10,15,20,25,30

30 comes in both time tables. 
it also does not cross 34

hope this helps you 

4 0
3 years ago
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