If the foreign exchange rate between the U.S. dollar and the Danish krone is 1:2.50, how many Danish kroner will you get for $5?

Find the value of x. O is the center of the circle. Round your answer to the nearest hundredth.

katrin [286]

$x=5.38$ . Correct option C)5.38

<u>Step-by-step explanation:</u>

Here we have , Find the value of x. O is the center of the circle. Round your answer to the nearest hundredth. picture is attached . Let's find out:

In the given figure , Let's draw a line from center to the point where cord of length 8 unit is touching the circle or intersecting or , where this chord finishes . This line is radius of circle and is denoted by x , So now we have a right angle triangle with dimensions as :

$Perpendicular =3.6\\Base=\frac{8}{2}=4\\Hypotenuse=x$

By Pythagoras Theorem ,

$Hypotenuse^2=Perpendicular^2+base^2$

⇒ $x^2=(3.6)^2+(4)^2$

⇒ $x=\sqrt{(3.6)^2+(4)^2}$

⇒ $x=\sqrt{12.96+16}$

⇒ $x=\sqrt{28.96}$

⇒ $x=5.38$

Therefore , $x=5.38$ . Correct option C)5.38

6 0

3 years ago

Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 6x(1/3) + 3x(4/3). You must justi

stealth61 [152]

Applying our power rule gets us our first derivative,

$\rm f'(x)=6\frac13x^{-2/3}+3\cdot\frac43x^{1/3}$

simplifying a little bit,

$\rm f'(x)=2x^{-2/3}+4x^{1/3}$

looking for critical points,

$\rm 0=2x^{-2/3}+4x^{1/3}$

We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.

$0=2x^{-2/3}\left(1+2x\right)$

When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.

Then apply your Zero-Factor Property,

$\rm 0=2x^{-2/3}\qquad\qquad\qquad 0=(1+2x)$

and solve for x in each case to find your critical points.

Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.

Let's come back to this,

$\rm f'(x)=2x^{-2/3}+4x^{1/3}$

and take our second derivative.

$\rm f''(x)=-\frac43x^{-5/3}+\frac43x^{-2/3}$

Looking for inflection points,

$\rm 0=-\frac43x^{-5/3}+\frac43x^{-2/3}$

Again, pulling out the smaller power of x, and fractional part,

$\rm 0=-\frac43x^{-5/3}\left(1-x\right)$

And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.

Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.

8 0

4 years ago

Roger gets $40 per day as wages and $4.50 as commission for every pair of shoes he sells in a day. His daily earnings goal is $1

il63 [147K]

I have no idea sorry

5 0

3 years ago

When x pounds of force is applied to one end of a lever that is L feet long, the resulting force y on the other end is

nadezda [96]

The question is worded poorly, but it looks like you have a lever in equilibrium, with a force x at a distance d from the fulcrum, and a force y at a distance L - d from the fulcrum. You already have the equilibrium formula for this situation:

xd = y(L - d)

If you know x, y, and d, you can solve for the length L.

3 0

3 years ago

(2) (x-7) (=3) (x+8)

Tasya [4]

Use the distributive property to get 2x - 14 = 3x + 24. Subtract 24 from both sides to get 2x - 38 = 3x. Subtract 2x from both sides to get -38 = x.

7 0

3 years ago

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