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3 years ago

Hong planted 16 tomato plants in his garden. What is the prime factorization of 16

Mathematics

2 answers:

anygoal [31]3 years ago

8 0

The prime factorization is 2^4

Lina20 [59]3 years ago

5 0

The prime factorization of 16 is 2^4

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Write the equation of the line that passes through the point(-6,12) and is parallel to the line y=8

Ugo [173]

Answer:

y = 12

Step-by-step explanation:

y = 8 is the equation of a horizontal line, parallel to the x- axis.

Thus a parallel line will also be horizontal.

The equation of a horizontal line is

y = c

where c is the y- coordinates of the points it passes through.

The line passes through (- 6, 12) with y- coordinate of 12, thus

y = 12 ← equation of parallel line

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3 years ago

Complete the pattern 444

astra-53 [7]

444..etc. etc. thats what i thought its simple really.

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3 years ago

How do I solve 2x+3y=7, x-y=1

Fittoniya [83]

<span>2x+3y=7
x-y=1

</span>2x+3y=7
x=1+y

2*(1+y)+3y=7
x=1+y

2+2y+3y=7
x=1+y

5y=5 |(:5)
x=1+y

y=1
x=2

CHECK:
2x+3y=7
x-y=1

2*2 + 3*1 = 7

x-y = 1
2-1 = 1

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3 years ago

Read 2 more answers

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

Evaluate q^2-5q+4, when q equals -2

Anastasy [175]

Answer:

Step-by-step explanation:

q^2 - 5q +4

Let q = -2

(-2)^2 - 5(-2) +4

4 + 10 +4

8 0

3 years ago