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2 years ago

7

Prove: The square of a number that is

1 answer:

Nikolay [14]2 years ago

3 0

Let 3<em>n</em> + 1 denote the "number" in question. The claim is that

(3<em>n</em> + 1)² = 3<em>m</em> + 1

for some integer <em>m</em>.

Now,

(3<em>n</em> + 1)² = (3<em>n</em>)² + 2 (3<em>n</em>) + 1²

… = 9<em>n</em>² + 6<em>n</em> + 1

… = 3<em>n</em> (3<em>n</em> + 2) + 1

… = 3<em>m</em> + 1

where we take <em>m</em> = <em>n</em> (3<em>n</em> + 2).

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Find m<br> A. 12<br><br> B. 22 <br><br> C. 52<br><br> D. 38

Nikolay [14]

C. 52

Work:

(5x - 22) + (4x + 4) = 90

9x - 18 = 90

9x = 108

x = 12

4(12) + 4 =

48 + 4 = 52

6 0

3 years ago

Is the circle inside the hexagon or hexagon inside the circle?

nexus9112 [7]

It could be either, we need more info. How big are each? If it doesn't say, then it's a trick question. Throw Schrodinger's cat at it or something.

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2 years ago

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dezoksy [38]

Answer:

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8 0

2 years ago

Please help me with my question!!

ollegr [7]

Answer:

10

Step-by-step explanation:

Since each ornament takes 1/4 of a piece of Bristol board we divide 2 1/2 by 1/4

2 1/2 = 5/2

(5/2)/(1/4) = 20/2 = 10

7 0

3 years ago

I am having trouble with this relative minimum of this equation.<br>

Norma-Jean [14]

Answer:

So the approximate relative minimum is (0.4,-58.5).

Step-by-step explanation:

Ok this is a calculus approach. You have to let me know if you want this done another way.

Here are some rules I'm going to use:

$(f+g)'=f'+g'$ (Sum rule)

$(cf)'=c(f)'$ (Constant multiple rule)

$(x^n)'=nx^{n-1}$ (Power rule)

$(c)'=0$ (Constant rule)

$(x)'=1$ (Slope of y=x is 1)

$y=4x^3+13x^2-12x-56$

$y'=(4x^3+13x^2-12x-56)'$

$y'=(4x^3)'+(13x^2)'-(12x)'-(56)'$

$y'=4(x^3)'+13(x^2)'-12(x)'-0$

$y'=4(3x^2)+13(2x^1)-12(1)$

$y'=12x^2+26x-12$

Now we set y' equal to 0 and solve for the critical numbers.

$12x^2+26x-12=0$

Divide both sides by 2:

$6x^2+13x-6=0$

Compaer $6x^2+13x-6=0$ to $ax^2+bx+c=0$ to determine the values for $a=6,b=13,c=-6$ .

$a=6$

$b=13$

$c=-6$

We are going to use the quadratic formula to solve for our critical numbers, x.

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-13 \pm \sqrt{13^2-4(6)(-6)}}{2(6)}$

$x=\frac{-13 \pm \sqrt{169+144}}{12}$

$x=\frac{-13 \pm \sqrt{313}}{12}$

Let's separate the choices:

$x=\frac{-13+\sqrt{313}}{12} \text{ or } \frac{-13-\sqrt{313}}{12}$

Let's approximate both of these:

$x=0.3909838 \text{ or } -2.5576505$ .

This is a cubic function with leading coefficient 4 and 4 is positive so we know the left and right behavior of the function. The left hand side goes to negative infinity while the right hand side goes to positive infinity. So the maximum is going to occur at the earlier x while the minimum will occur at the later x.

The relative maximum is at approximately -2.5576505.

So the relative minimum is at approximate 0.3909838.

We could also verify this with more calculus of course.

Let's find the second derivative.

$f(x)=4x^3+13x^2-12x-56$

$f'(x)=12x^2+26x-12$

$f''(x)=24x+26$

So if f''(a) is positive then we have a minimum at x=a.

If f''(a) is negative then we have a maximum at x=a.

Rounding to nearest tenths here: x=-2.6 and x=.4

Let's see what f'' gives us at both of these x's.

$24(-2.6)+25$

$-37.5$

So we have a maximum at x=-2.6.

$24(.4)+25$

$9.6+25$

$34.6$

So we have a minimum at x=.4.

Now let's find the corresponding y-value for our relative minimum point since that would complete your question.

We are going to use the equation that relates x and y.

I'm going to use 0.3909838 instead of .4 just so we can be closer to the correct y value.

$y=4(0.3909838)^3+13(0.3909838)^2-12(0.3909838)-56$

I'm shoving this into a calculator:

$y=-58.4654411$

So the approximate relative minimum is (0.4,-58.5).

If you graph $y=4x^3+13x^2-12x-56$ you should see the graph taking a dip at this point.

3 0

3 years ago