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3 years ago

-1/2 + 2/3 ????????????????????????????????

Mathematics

1 answer:

Free_Kalibri [48]3 years ago

6 0

Answer:

2/6 or simpler 1/3

Step-by-step explanation:

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Hernando and Rachel are factoring 2mp-6p+27-9m. Is either of them correct? Explain your reasoning,

vichka [17]

Answer:

See Explanation

Step-by-step explanation:

The question is incomplete, as Hernando and Rachel's solution are not provided. So, I will just solve the question directly.

Given

$2mp-6p+27-9m$

Required

Factor

$2mp-6p+27-9m$

Group into 2

$2mp-6p+27-9m = [2mp-6p]+[27-9m]$

Factor each group

$2mp-6p+27-9m = 2p[m-3]+9[3-m]$

Rewrite 3 - m as -(m-3)

So, we have:

$2mp-6p+27-9m = 2p[m-3]-9[m-3]$

Factor out m - 3

$2mp-6p+27-9m = [2p-9][m-3]$

3 0

3 years ago

I need a little help -

OverLord2011 [107]

Answer:

(6x6x6x6) x (6x6x6)

Step-by-step explanation:

6 to the power of four is just 6 multiplied to itself 4 times.

6 to the power of three is just 6 multiplied to itself 3 times.

So, you would be multiplying 6, 7 times.

You could check it with a calculator. - 6^4 x 6^3 = 279936.

6^7 = 279936.

8 0

1 year ago

HELP

kati45 [8]

Slope: 1/5

Y-intercept: 0

Equation: Y= 5x

5 0

3 years ago

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

2 years ago

If 1/3tr=uw, then __?

Alisiya [41]

Answer : If you mean this: 1/(3TR) = UV

Then the answer is : 1 = 3TRUV

3 0

3 years ago