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3 years ago

Someone please answer this for me ASAP

Mathematics

2 answers:

gogolik [260]3 years ago

8 0

The answers are

31
7
1
-5

i hope this helped. have a good day

NISA [10]3 years ago

5 0

-5, 31
-1, 7
0, 1
1, -5

The best way to find the answer to this question is to type in the equation into a graphing calculator under “y=“ then press the graph/ table button!

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Whereas the basic formula for the area shape is length x width the basic formula for volume is length x width c height

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2 years ago

A man in a tower spots three hot air balloons flying at the same altitude. Balloon

anyanavicka [17]

Answer:

90/4= 12.9

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Step-by-step explanation:

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2 years ago

3 5⁄7 + 1 9⁄14 = What is the answer if the answer is wrong I will come and haunt you

slega [8]

Answer:

$5 \frac{5}{14}$

Step-by-step explanation:

$3 \frac{5}{7} + 1 \frac{9}{14}$

Turn them into improper fractions

$\frac{26}{7} + \frac{23}{14}$

$\frac{2}{2} \times \frac{26}{7} + \frac{23}{14}$

$\frac{52 + 23}{14}$

$\frac{75}{14}$

Turn it into a mixed fraction

$5 \frac{5}{14}$

7 0

2 years ago

Please help asap! thank you. :)

Black_prince [1.1K]

X<22
x>-7
x<13
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x>-39
i hope this helps

5 0

3 years ago

At time t=0 sec, a tank contains 15 oz of salt dissolved in 50 gallons of water. Then brine containing 88oz of salt per gallon o

maria [59]

Answer:

a) $s(t) =400- 385 e^{-\frac{1}{10} t}$

b) $s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397$

Step-by-step explanation:

Part a

Assuming that the original concentration of salt is 8 oz/gal and that the rate of in is equal to the rate out = 5 gal/min.

For this case we know that the rate of change can be expressed on this way:

$Rate change = In-Out$

And we can name the rate of change as $\frac{ds}{dt}=rate change$

And our variable s would represent the amount of salt for any time t.

We know that the brine containing 8oz/gal and the rate in is equal to 5 gal/min and this value is equal to the rate out.

For the concentration out we can assume that is $\frac{s}{50gal}$

And now we can find the expression for the amount of salt after time t like this:

$\frac{dS}{dt}= 8 \frac{oz}{gal}(5\frac{gal}{min}) -\frac{s}{50gal} 5 \frac{gal}{min} =40\frac{oz}{min}- \frac{s}{10} \frac{oz}{min}$

And we have this differential equation:

$\frac{dS}{dt} +\frac{1}{10} s = 40$

With the initial conditions y(0)=15 oz

As we can see we have a linear differential equation so in order to solve it we need to find first the integrating factor given by:

$\mu = e^{\int \frac{1}{10} dt }= e^{\frac{1}{10} t}$

And then in order to solve the differential equation we need to multiply with the integrating factor like this:

$e^{\frac{1}{10} t} s = \int 40 e^{\frac{1}{10} t} dt$

$e^{\frac{1}{10} t} s = 400 e^{\frac{1}{10} t} +C$

Now we can divide both sides by $e^{\frac{1}{25} t}$ and we got:

$s(t) =400 + C e^{-\frac{1}{10} t}$

Now we can apply the initial condition in order to solve for the constant C like this:

$15 = 400+C$

$C=-385$

And then our function would be given by:

$s(t) =400- 385 e^{-\frac{1}{10} t}$

Part b

For this case we just need to replace t =25 and see what we got for the value of the concentration:

$s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397$

7 0

3 years ago