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Advocard [28]
3 years ago
9

What is the range of -.5x2+3x-2

Mathematics
2 answers:
Helga [31]3 years ago
6 0
Ur answer is 8x to ur problem and if u need mr to explain i will be glad to
RoseWind [281]3 years ago
6 0

Answer:Your answer will be 8x

Step-by-step explanation:

If your confused just tell me and I will explain it

You might be interested in
Find the slope (m) and the y intercept please show work
gregori [183]

Answer:

m=2/5

b=3.8

Step-by-step explanation:

To find slope you need to use slope formula which is (y2-y1)/(x2-x1) so you need two points from the table and plug them into this. I'll use the first two points (-2, 3) and (-1, 5)

(5-3)/(3--2)

2/5

our slope is 2/5(m)

now to find the y int(b)

we have out slope intercept formula which is y=mx+b

plug in our slope

y=2/5x+b

now take a random point from our tabell and apply it to our equation I'll use (-2, 3) again.

3=2/5(-2)+b

3=-4/5+b

3=-0.8+b

add 0.8 to each side

3.8=b

b=3.8

4 0
3 years ago
Please help me I need help for this
amid [387]

Answer:

the greatest common factor is 4a

5 0
3 years ago
Read 2 more answers
Find three consecutive odd integers such that the sum of the​ first, two times the second and three times the third is 82.
tigry1 [53]
I think the answer is (A)
4 0
3 years ago
Read 2 more answers
The population of mosquitoes in a certain area increases at a rate proportional to the current population, and in the absence of
Alexandra [31]

Answer:

Population of mosquitoes in the area at any time t is:

P(t) =504,943.26  -104,943.26e^{0.693t}

Step-by-step explanation:

assume population at any time t = P(t)

population increases at a rate proportional to the current population:

⇒dP/dt ∝ P

 \implies \frac{dP}{dt} =kP----(1)

where k is constant rate at which population is doubled

solving (1)

ln|P(t)|=kt +C\\P(t)= e^{kt+C}\\P(t)=Ce^{kt}

t=0\\P(0) = P_{o}\\\implies C= P_{o}\\P(t) =P_{o}e^{kt}\\ ---- (2)

initial population = 400,000

population is doubled every week

                                                 ⇒P(1)=2P(0)

Using (2)

                                 P_{o}e^{k(1)} = 2P_{o} e^{k(0)}\\

                                            e^{k} =2\\k=ln|2|\\

In presence of predators amount is decreased by 50,000 per day

Then amount decreased per week = 350,000

In this case (1) becomes

\frac{dP}{dt}=kP-350,000\\\frac{dP}{dt} - kP=-350,000\\ ---(3)

solving (3) by calculating integrating factor

                                          I.F=e^{\int-k dt}

Multiplying I.F with all terms of (3)

e^{-kt}\frac{dP}{dt} - ke^{-kt}P =-350,000 e^{-kt}\\\frac{d}{dt}(e^{-kt}P) =  -350,000 e^{-kt}

Integrating w.r.to t

                         e^{-kt}P(t)= \frac{350,000e^{-kt}}{k} +C

                         P(t) =\frac{350,000}{k} +Ce^{kt}\\

                                          k=ln|2| =0.693

                          P(t) =504,943.26 + Ce^{0.693t}\\

at t=0

                        P(0) =504,943.26 + Ce^{0.693(0)}

                        400,000 =504,943.26 + C

                           C = -104,943.26

So, population of mosquitoes in the area at any time t is

                  P(t) =504,943.26  -104,943.26e^{0.693t}

6 0
3 years ago
What is the common denominator of (5/x^2-4) - (2/x+2) in the complex fraction (2/x-2) - (3/x^2-4)/(5/x^2-4) - (2/x+2)
PIT_PIT [208]

9514 1404 393

Answer:

  • common denominator: (x² -4)
  • simplified complex fraction: (2x +1)/(9 -2x)

Step-by-step explanation:

It is helpful to remember the factoring of the difference of squares:

  a² -b² = (a -b)(a +b)

__

Your denominator of (x² -4) factors as (x -2)(x +2). You will note that one of these factors is the same as the denominator in the other fraction.

It looks like you want to simplify ...

  \dfrac{\left(\dfrac{2}{x-2}-\dfrac{3}{x^2-4}\right)}{\left(\dfrac{5}{x^2-4}-\dfrac{2}{x+2}\right)}=\dfrac{\left(\dfrac{2(x+2)}{(x-2)(x+2)}-\dfrac{3}{(x-2)(x+2)}\right)}{\left(\dfrac{5}{(x-2)(x+2)}-\dfrac{2(x-2)}{(x-2)(x+2)}\right)}\\\\=\dfrac{2(x+2)-3}{5-2(x-2)}=\boxed{\dfrac{2x+1}{9-2x}}

3 0
3 years ago
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