All categories

3 years ago

How can I do number 25?

Mathematics

2 answers:

vlada-n [284]3 years ago

4 0

Perpendicular lines looks like a "+" the point is in the middle at the point of intersection.

Delicious77 [7]3 years ago

3 0

+
^ that's your answer

You might be interested in

Help meeeeeeeeeeeeeeeeeeeeeee !!!!!

stiv31 [10]

Answer:

i think it is the 2nd one and the 5th one.

Step-by-step explanation:

hope this helped!

p.s it would be cool if you gave me brainliest

6 0

3 years ago

I really need help????<br> Plz answer quickly

Julli [10]

Answer:

3 7 8 2 6 4 1 10 3 5 9 the answer

6 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

1 year ago

Andy wrote the equation of a line that has a slope of Step 2: y plus 2 equals StartFraction 3 Over 4 EndFraction x minus StartFr

nlexa [21]

answer is C

He made an error in Step 3. He should have subtracted 2 from both sides.

4 0

3 years ago

Read 2 more answers

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

liubo4ka [24]

Answer:

1) 20.9

2) 896

3) 21

Step-by-step explanation:

1) 5.6÷2^3+(12.75+7.45)

---> 12.75 + 7.45 = 20.2

$5.6$ ÷ $2^3 + 20.2$

--> Simplify 2^3 to 8

$5.6$ ÷ 8 + 20.2

--> 5.6 ÷ 8 = 0.7

$0.7+20.2$

--> Simplify

$20.9$

2) 4^3 × (0.6 +3.6) ÷ 0.3

---> 0.6 + 3.6 = 4.2

4^3 * 4.2 ÷ 0.3

---> 4^3 = 64

64 * 4.2 ÷ 0.3

--> 64 * 4.2 = 268.8

268.8 ÷ 0.3

--> 268.8 ÷ 0.3 = 896

896

3) 2^4 + (2.75 +1.75) ÷ 0.9

--> 2.75 + 1.75 = 4.5

2^4 + 4.5 ÷ 0.9

--> 2^4 = 16

16 + 4.5 ÷ 0.9

--> 4.5 ÷ 0.89 = 5

16 + 5

--> Simplify

= 21

8 0

2 years ago