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3 years ago

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