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3 years ago

Which equation matches the graph of the greatest integer function given below?? help pleaze

Mathematics

2 answers:

Alex787 [66]3 years ago

4 0

The correct answer is C

aleksandr82 [10.1K]3 years ago

4 0

Answer:

The equation which matches the graph of the greatest integer function given is:

C) $y=[x]+3$

Step-by-step explanation:

From the graph we may observe that:

when x=0

then the value of the function is: 3

Hence, we will put the value of x in each of the given options and see what holds true.

$y=[x]-3$

when x=0 we have:

$y=[0]-3\\\\i.e.\\\\y=-3\neq 3$

Hence, option: A is incorrect.

$y=[x]+1$

when x=0 we have:

$y=[0]+1\\\\i.e.\\\\y=1\neq 3$

Hence, option: B is incorrect.

$y=[x]-1$

when x=0 we have:

$y=[0]-1\\\\i.e.\\\\y=-1\neq 3$

Hence, option: D is incorrect.

$y=[x]+3$

when x=0 we have:

$y=[0]+3\\\\i.e.\\\\y=3$

Similarly all the other values hold for this function.

and the graph of this function matches the given graph.

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True or False: Only like terms can be combined. A True B False

mina [271]

Answer:

True

Step-by-step explanation:

Like terms like 4x-3x the x term. is a like term

6-4x can be subtracted because they aren't the same.

High Hopes^^

Barry-

3 0

3 years ago

Read 2 more answers

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows

4 0

3 years ago

pentagon [3]

Answer:

cant see picture but it is an easy question based on equilateral that has 3 identical angles and is congruent. A Scalene has odd sides and is not congruent and an isosceles has two sides identical. You can also get two type of right angled triangle only 45 45 plus 90 for right angle would be be congruent

Step-by-step explanation:

3 0

2 years ago

Write a quadratic equation with the given roots. Write the equation in the form of ax^2+bx+c=0 where a b and c are integers

Bas_tet [7]

You haven't provided the required roots, but I can tell you how to do this kind of exercises in general.

If the $x^2$ coefficient is 1, i.e. the equation is written like $x^2+bx+c=0$ , then you can say the following about the coefficients b and c:

$b$ is the opposite of the sum of the roots
$c$ is the multiplication of the roots.

So, for example, if we want an equation whose roots are 4 and -2, we have:

$4+(-2) = 4-2 = 2 \implies b = -2$
$4 \cdot (-2) = -8 \implies c = -8$

So, the equation is $x^2-2x-8=0$

If your roots are rational, you can work like this: suppose you want an equation with roots 3/4 and 1/2. You have:

$\dfrac{3}{4}+\dfrac{1}{2} = \dfrac{3}{4}+\dfrac{2}{4} = \dfrac{5}{4} \implies b = -\dfrac{5}{4}$
$\dfrac{3}{4} \cdot \dfrac{1}{2} = \dfrac{3}{8} \implies c = \dfrac{3}{8}$