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cluponka [151]
3 years ago
6

I need help asap pllz help !!​

Mathematics
1 answer:
swat323 years ago
6 0

Answer:

14/28

Step-by-step explanation:

14/28 reduces down to 1/2

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aniked [119]

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10v*3 =

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Step-by-step explanation:

That's the answer

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Composite function is the result of using one function as the input for another function.
ipn [44]

Answer:

True

Step-by-step explanation:

That is the definition of composite function

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3 years ago
Can anyone help me? My teacher just recently taught me this today, be he never explained much. The question is in the photo
IceJOKER [234]

You've got five different problems in this photo ... four on top and the word problem on the bottom ... and they're all exactly the same thing:  Taking two points and finding the slope of the line that goes through them.

In every case, the procedure is the same.
If the two points are  (x₁ , y₁)  and  (x₂ , y₂) , then
the slope of the line that goes through them is

                          Slope  =  (y₂ - y₁) / (x₂ - x₁) .

This is important, and you should memorize it.

#1).  (8, 10)  and  (-7, 14)

         Slope  =  (14 - 10) / (-7 - 8)  =  4 / -15

#2).  (-3, 1)  and  (-17, 2)

         Slope  =  (2 - 1) / (-17 -  -3)  =  (2 - 1) / (-17 + 3)  =  1 / -14

#3).  (-20, -4)  and  (-12, -10)

         Slope  =  [ -10 - (-4) ] / [ -12 - (-20) ]

=========================================

The word problem:

This question only gives you one point on the graph,
and then it wants to know what's the slope ?
What are you going to do for another point ?

A "proportional relationship" always passes through the origin,
so another point on the line is  (0, 0) .

Now you have two points on THAT line too, and you can easily
find its slope.

4 0
3 years ago
A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th
Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let S(t) be the amount of sugar in the tank at time t. Then S(0)=25.

Sugar is added to the tank at a rate of <em>P</em> lb/min, and removed at a rate of

\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}

Separate variables, integrate, and solve for <em>S</em>.

\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt

\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt

-100\ln\left|P-\dfrac S{100}\right|=t+C

\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t

P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}

\dfrac S{100}=P-Ce^{-100t}

S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}

Use the initial value to solve for <em>C</em> :

S(0)=25\implies 25=100P-C\implies C=100P-25

\implies S(t)=100P-(100P-25)e^{-100t}

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick <em>P</em> such that

S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}

5 0
3 years ago
3x(x-2)-x^2+4=0<br> Help
antoniya [11.8K]

Answer:

x=2,1 solving with quadratic formula

Step-by-step explanation:

3 0
3 years ago
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