Solving quadratic equations by factoring m(m-3)=0

1 answer:

6 0

Because it has already been factored, we can solve it straight away. Now, since m is a variable, we yield the same result if m = 0 or m = 3 because we have two components to this equation.

We can either have m = 0 or m - 3 = 0
Hence, m is either 0 or 3

You might be interested in

Can I plss have an answer on this i need it ASAP plss

nordsb [41]

WHAT DOES THIS MEAN???... i needa think abt this one

4 0

2 years ago

Find w(4) for the function w(x) = 3x + 7.

monitta

W(4) = 3(4) + 7
w(4) = 12 + 7
w(4) = 19

7 0

3 years ago

Read 2 more answers

Factor the polynomial, given that x=-3 is a zero

GREYUIT [131]

if x-2 is a factor of polynomial, f (×) ,then it means we can find the zeros or the solutions of the polynomial by putting for (×-2)=0 . Because the zeros or roots of a polynomial is found by putting f (×)=0 , where x is a factor of the polynomial

3 0

2 years ago

A restaurant offers a "Create Your Own Pizza" menu option. There are 33 types of meat, 66 types of vegetables, and 33 kinds of c

OleMash [197]

Answer:

Step-by-step explanation:

You have to do the number of options in each category by itsself

5 0

3 years ago

A 500500-gallon tank initially contains 200200 gallons of brine containing 100100 pounds of dissolved salt. brine containing 11

mixer [17]

Let $A(t)$ denote the amount of salt in the tank at time $t$ . We're given that the tank initially holds $A(0)=100$ lbs of salt.

The rate at which salt flows in and out of the tank is given by the relation

$\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}$
$\implies A'(t)+\dfrac{11}{200+33t}A(t)=484$

Find the integrating factor:

$\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}$

Distribute $\mu(t)$ along both sides of the ODE:

$(200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}$
$\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}$
$A(t)=484\displaystyle\int(200+33t)^{-1/3}\,\mathrm dt$
$A(t)=22(200+33t)^{2/3}+C$

Since $A(0)=100$ , we get

$100=22(200)^{2/3}+C\implies C\approx-652.39$

so that the particular solution for $A(t)$ is

$A(t)=22(200+33t)^{2/3}-652.39$

The tank becomes full when the volume of solution in the tank at time $t$ is the same as the total volume of the tank:

$200+(44-11)t=500\implies 33t=300\implies t\approx9.09$

at which point the amount of salt in the solution would be

$A(9.09)\approx733.47\text{ lb}$

4 0

3 years ago