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3 years ago

Can anyone help? I will mark you as brainliest!

Mathematics

1 answer:

Rudik [331]3 years ago

3 0

Answer: Part A is - the highest frequency was 5

The total number of the frequency is 24

The range is 1

I gave you the best answer that I have. I did my best. Good luck! :)

Step-by-step explanation:

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Addition (must show work) can someone solve 1. m + 13 =29 2. z + x =34 <br><br> 3. 87 + p = 136

azamat

Greetings ! Your Answer Is Below.

Answer;

1. m = 16

2. x = -z + 34 or z = -x + 34

3. p = 49

Step-by-step explanation;

1. m + 13 - 13 = 29 - 13 | Subtract 13 from both sides ( m = 16 )

2. - z + 34 | Add -z to both of the sides ( x = - z + 34 )

Or - x + 34 | Add -x to both of the sides ( z = - x + 34 )

3. 136 - 87 = 49 | Subtract 87 from both sides ( p = 49 )

[ Hope This Helped ! ]

3 0

3 years ago

Expand logarithms equation

Bess [88]

Answer:

$log_2(x) + log_2(y) + log_2(z)$

Step-by-step explanation:

$5log_2(xyz)^{1/5}\\\\5log_2(x)^{1/5} + 5log_2(y)^{1/5} + 5log_2(z)^{1/5}\\\\log_2(x) + log_2(y) + log_2(z)$

4 0

3 years ago

1,840 mL = ____ L<br><br> WHAT DOES L EQUAL PLZ HELP!!!

sweet-ann [11.9K]

It equals 1.84 liters

6 0

3 years ago

Read 2 more answers

given the rule s=t+3 and the starting number 0, create a table to show the first 5 terms in the sequence graph the resulting ord

IRINA_888 [86]

0= 3
1= 4
2=5
3=6
4=7
5=8
6=9
7=10

8 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