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Mkey [24]
3 years ago
15

For which function is the domain

Mathematics
1 answer:
Anettt [7]3 years ago
3 0

Answer:

D. y=\sqrt{x}

Step-by-step explanation:

A square root can't be evaluated if the number we want to find the square of  is nagative.

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What is the maximum number of possible extreme values for the function,<br> f(x) = х2 -7x – 6?
timama [110]

Answer:

Only one extreme value of f(x) is possible.

Step-by-step explanation:

We are given the quadratic function of independent variable x which is f(x) = x² - 7x - 6 ......(1)

Now. the condition for extreme values of f(x) is \frac{df(x)}{dx} =0

Hence, differentiating both sides of equation (1) with respect to x, we get

\frac{df(x)}{dx} = 2x -7 = 0

⇒ x = 3.5.

So there is only one value of x for which f(x) has extreme value which is x = 3.5.

Therefore, only one extreme value of the given function is possible. (Answer)

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3 years ago
Solve this please...
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Which quadrilaterals always have two pairs of congruent opposite angles?
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2 years ago
The general solution to the second-order differential equation y′′+10y=0 is in the form y(x)=c1cosβx+c2sinβx. Find the value of
allsm [11]

Answer:β=√10 or 3.16 (rounded to 2 decimal places)

Step-by-step explanation:

To find the value of β :

  • we will differentiate the y(x) equation twice to get a second order differential equation.
  • We compare our second order differential equation with the Second order differential equation specified in the problem to get the value of β

y(x)=c1cosβx+c2sinβx

we use the derivative of a sum rule to differentiate since we have an addition sign in our equation.

Also when differentiating Cosβx and Sinβx we should note that this involves function of a function. so we will differentiate  βx in each case and multiply with the differential of c1cosx and c2sinx respectively.

lastly the differential of sinx= cosx and for cosx = -sinx.

Knowing all these we can proceed to solving the problem.

y=c1cosβx+c2sinβx

y'= β×c1×-sinβx+β×c2×cosβx  

y'=-c1βsinβx+c2βcosβx

y''=β×-c1β×cosβx + (β×c2β×-sinβx)

y''= -c1β²cosβx -c2β²sinβx

factorize -β²

y''= -β²(c1cosβx +c2sinβx)

y(x)=c1cosβx+c2sinβx

therefore y'' = -β²y

y''+β²y=0

now we compare this with the second order D.E provided in the question

y''+10y=0

this means that β²y=10y

β²=10

B=√10  or 3.16(2 d.p)

5 0
3 years ago
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