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Is 2 the solution to x to the third minus x minus 6 equals 0

Natali [406]

X actually equals the cube root of 6
${x}^{3} - 6 = 0 \\ {x}^{3} = 6 \\ x = \sqrt[3]{6}$

5 0

3 years ago

Circle 1: center (8, 5) and radius 6

PolarNik [594]

We have that
Circle 1: center (8, 5) and radius 6
Circle 2: center (−2, 1) and radius 2

we know that
the equation of a circle is
(x-h)²+(y-k)²=r²
for the circle 1---------> (x-8)²+(y-5)²=36
for the circle 2---------> (x+2)²+(y-1)²=4

using a graph tool
see the attached figure

Part A)What transformations can be applied to Circle 1 to prove that the circles are similar?

we know that
r1/r2---------> 6/2------> 3

to prove that the circle 1 and circle 2 are similar, the radius of circle 1 must be divided by 3 and translate the center of the circle 1 (10) units to the left and (4) units down

the answer part A) is
the radius of circle 1 must be divided by 3 and translate the center of the circle 1 (10) units to the left and (4) units down

Part B) What scale factor does the dilation from Circle 1 to Circle 2 have?

the answer Part B) is
the scale factor is (3/1)

6 0

3 years ago

Prove for any positive integer n, n^3 +11n is a multiple of 6

suter [353]

There are probably other ways to approach this, but I'll focus on a proof by induction.

The base case is that n = 1. Plugging this into the expression gets us

n^3+11n = 1^3+11(1) = 1+11 = 12

which is a multiple of 6. So that takes care of the base case.

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Now for the inductive step, which is often a tricky thing to grasp if you're not used to it. I recommend keeping at practice to get better familiar with these types of proofs.

The idea is this: assume that k^3+11k is a multiple of 6 for some integer k > 1

Based on that assumption, we need to prove that (k+1)^3+11(k+1) is also a multiple of 6. Note how I've replaced every k with k+1. This is the next value up after k.

If we can show that the (k+1)th case works, based on the assumption, then we've effectively wrapped up the inductive proof. Think of it like a chain of dominoes. One knocks over the other to take care of every case (aka every positive integer n)

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Let's do a bit of algebra to say

(k+1)^3+11(k+1)

(k^3+3k^2+3k+1) + 11(k+1)

k^3+3k^2+3k+1+11k+11

(k^3+11k) + (3k^2+3k+12)

(k^3+11k) + 3(k^2+k+4)

At this point, we have the k^3+11k as the first group while we have 3(k^2+k+4) as the second group. We already know that k^3+11k is a multiple of 6, so we don't need to worry about it. We just need to show that 3(k^2+k+4) is also a multiple of 6. This means we need to show k^2+k+4 is a multiple of 2, i.e. it's even.

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If k is even, then k = 2m for some integer m

That means k^2+k+4 = (2m)^2+(2m)+4 = 4m^2+2m+4 = 2(m^2+m+2)

We can see that if k is even, then k^2+k+4 is also even.

If k is odd, then k = 2m+1 and

k^2+k+4 = (2m+1)^2+(2m+1)+4 = 4m^2+4m+1+2m+1+4 = 2(2m^2+3m+3)

That shows k^2+k+4 is even when k is odd.

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In short, the last section shows that k^2+k+4 is always even for any integer

That then points to 3(k^2+k+4) being a multiple of 6

Which then further points to (k^3+11k) + 3(k^2+k+4) being a multiple of 6

It's a lot of work, but we've shown that (k+1)^3+11(k+1) is a multiple of 6 based on the assumption that k^3+11k is a multiple of 6.

This concludes the inductive step and overall the proof is done by this point.

6 0

3 years ago

Read 2 more answers

Trigonometry please help! work need to be shown

padilas [110]

Answer:

x = 5.56

Step-by-step explanation:

Apply trigonometric function to find x

Reference angle = 46°

Hypotenuse = 8

Adjacent length = x

Apply CAH:

Cos 46 = Adj/Hyp

Cos 46 = x/8

8 × Cos 46 = x

5.55726696 = x

x = 5.56 (nearest hundredth)

7 0

3 years ago

How may square roots does it to take to get a square root of the square root?

Xelga [282]

If you want to take a square root of a square root you will need 2 squares roots,

For example, i want the square root of the square root of 16.

I'll take it first: √16 = 4 and again, √4 = 2

So, the square root of the square root of 16 is 2.

But you can also make a root with index 4.

$\sqrt[4]{16}=2$

Because if you do 2.2.2.2 you will take 16.

3 0

4 years ago