Which is least like the others: rotator, deified, race car, alright

Answers

Answer 1

Answer: The correct answer is "alright". This is the only word in this list that is not a palindrome (a word that is the same forward and backward).

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Circle Q has the central angle ∠AQB with a measure of 55°. Which of the following option is not true? AB is a minor arc
the measure of AB=110 degrees
The center of the circle point is Q
The measure of AB=55 degrees

Answers

Answer with explanation:

It is given about a Circle centered at Q, having measure of Central Angle AQB =55°

(A). Which says that AB is minor arc.Because of measure of central angle is , 0°<Central Angle <180°.

(B). The Measure of arc AB ≠110°, Since ,∠AQB=55°, Nor Reflex (∠AQB)≠110°.

(C). Yes, it is true statement , that center of this circle is at point Q.

(D). The Measure of minor arc,AB making angle with the Point Q is equal to =55° is certainly true Statement.

Among , the four statement , Second assertion is false which states that,the measure of AB=110 degrees .

The measure of AB is 110 degrees

A sequence is defined recursively by f(1)=16 and f(n)= f(n-1)+2n. Find f(4)

Answers

Hi,

f(1)=16
f(2)=f(1)+2*2=16+4=20
f(3)=f(2)+2*3=20+6=26
f(4)=f(3)+2*4=26+8=34

Find the value of y.

Answers

You're working with two right triangles here.

The big one is the whole drawing. It's lying flat on its hypotenuse.
One side of the big triangle ... the left side ... has no label.
We'll be working with that side. Let's call it ' S ' .

The smaller triangle is the left piece of the big one.
Its height is 'y', and its base is 4. ' S ' is its hypotenuse.

Go to your toolbox and pull out the little box marked "Pythagoras".
Inside that box is the equation you need in order to work on this
problem, or any problem with right triangles. The equation is ...

   (hypotenuse)² = (length of one leg)² + (length of the other leg)² .

In the big triangle, 'x' is one leg, and the hypotenuse is 16 .

   (16)² = x² + S²

   256 = x² + S²

Subtract x² from each side:    S² = 256 - x²

Take the square root of each side: S = √(256 - x²)
___________________________________

Now look at the small right triangle.

One leg is 4. The other leg is 'y'. The hypotenuse is ' S '.

S² = (4)² + y²

Subtract 16 from each side: y² = S² - 16

Writer in the value of S² that
we worked out before: y² = (256 - x²) - 16 .

   y² = 240 - x²

Take the square root
of each side: y = √(240 - x²) .

That's what you're supposed to find. So we're done.
_________________________________________

Oh phooey !   Sometimes I am so stupid.

Forget almost everything I did up above, and
let's back up several steps:

==> Cut the picture along the line marked 'y', and throw away
   the left piece, with the '4' on the bottom. We don't need it.

-- Now you have a right triangle that I didn't even notice before.
   One leg is 12. The other leg is 'y'. The hypotenuse is 'x'.

Pythagoras says:    (x²) = (12²) + (y²)

Subtract 12² from each side: y² = x² - 144

Take the square root of each side: y = √(x² - 144) .

We know that the shortest leg of the largest right triangle (hypotenuse length 16) has a length of the square root of (y^2 + 16) through the pythagoras theorem. We also know that x is equal to the square root of (144 + y^2) as well. Finally, we know that 256 = (y^2 + 16) + (y^2 + 144), since we use the pythagoras theorem one last time and square each of the values we have got for the two legs of the largest right triangle. Therefore, 256 = 2y^2 + 160, 96 = 2y^2, 48 = y^2, y= sqrt48 = 4sqrt3 = $4√(3)$ .