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Citrus2011 [14]
2 years ago
6

I accidentally clicked on B. I don’t understand please help

Mathematics
1 answer:
Pavel [41]2 years ago
4 0

Answer:

{-3,2,5}

hope this helps :) plz brainliest if can

Step-by-step explanation:

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It is now 8:59 a.m. when the bell rings at 9:01 am. Damian will be late for science class for the 3rd time this week. He must ge
kvv77 [185]

Answer:

1: 10 seconds   2: 4 seconds  3: 12 seconds

Step-by-step explanation:

35.0/3.5= 10

4.80/1.20= 4

60.0/5.00= 12

4 0
2 years ago
What is the value of z in the equation 2z + 6 = –4? (5 points) <br> 5 <br> 1 <br> –1 <br> –5
PtichkaEL [24]
The answer is -5. Hope this helps.
6 0
3 years ago
Read 2 more answers
manuel has $15 to spend.on apps and music. Apps cost $2 each and songs cost $1 each. If x represents the number of apps purchase
strojnjashka [21]

Answer:

2x + y = 15

Step-by-step explanation:

Total amount with Manuel = $15

Cost of apps = $2

Cost of song = $1

x = number of apps purchased

y = number of songs purchased

Cost of apps × number of apps purchased + Cost of song × number of songs purchased = Total amount with Manuel

2*x + 1*y = 15

2x + y = 15

The linear equation that represents the combination of apps and songs manuel can purchase is 2x + y = 15

6 0
3 years ago
The ratio of boys to girls in Ms. smith’s class is 3 to 4. If there are 12 boys, how many girls are there?
Lorico [155]

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

                  Hello!

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

❖ There will be 16 girls.

The relationship is x4. If we multiply 3 x 4, we get 12. What we do to one number we do to the other so 4 x 4 = 16.

~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ

4 0
3 years ago
A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and
goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}

Solve for S(t):

\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}

e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}

The left side is the derivative of a product:

\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}

Integrate both sides:

e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt

e^{t/800}S(t)=64e^{t/800}+C

S(t)=64+Ce^{-t/800}

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

0=64+C\impleis C=-64

and so (b) the amount of sugar in the tank at time t is

S(t)=64\left(1-e^{-t/800}\right)

As t\to\infty, the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

7 0
3 years ago
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