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there are 10 ways that 3 consecutive people can be chosen.

There are 10* 6 ways that 2 (but not 3) consecustive people can be chosen

So that is  70 ways that at least 2 consecutive people are chosen.

There are 10C3 = 120 ways that any 3 people can be chosen

so maybe  70/120 ways.

That is my best guess.

A parabola with equation y=ax^2+bx+c has a vertical line of symmetry at x=2 and goes through the two points (1, 1) and (4, -7). The quadratic ax^2+bx+c has two real roots. The greater root is sqrt(n)+2. What is n?

The axis of symetry is just   -b/2a 

so   -b/2a = 2

-b = 4a

b=-2a

So the equation so far becomes       y=ax^2 -2ax  +c

Sub in your two points and solve simulataneously and what do you get for a and c

Can you take it from here?

4/5 meters in 8 minutes

is

8 minutes for  4/5 of a meter

8*5/4  minutes for  4/5  *  5/4 metres.

you can finish it.

 Find the area of triangle OBC .

Hello Guest!

The triangle is symmetrical in the X-Y coordinate system with BC on the ordinate axis.

\(A(0,\sqrt{5^2-3.5^2})= A(0,\sqrt{12.75})\\ B(-3.5,\ 0)\\ C(3.5,\ 0)\\ M(\frac{7}{4},\frac{\sqrt{12.75}}{2})\ [center\ \overline{AC}]\\ m_{AC(x)}=-\frac{\sqrt{12.75} }{3.5}\\ m=\frac{3.5}{\sqrt{12.75}}\)

\(MO(x)=m_{}(x-x_M)+y_M\\ MO(x)=\frac{{3.5}}{\sqrt{12.75}}(x-\frac{7}{4})+\frac{3.5707}{2}\\ h=MO(0)=\frac{{3.5}}{\sqrt{12.75}}(0-\frac{7}{4})+\frac{\sqrt{12.75}}{2}\\ h=0.0700147\\ A_{OBC}=\dfrac{0,0700147\cdot 7}{2}\\ \color{blue}A_{OBC}=0.1225245\\ Finally\ solved!\\ (The\ circular\ function:\ f(x)=\sqrt{3.5^2+0.0700147^2-x^2}+0.0700147)\)

  !

Since the pizza order is larger than $75, it qualifies for the coupon.

In order to find the percent they saved, you take 12 and divide it by 96.

\(\frac{12}{96}\) = 0.125

Convert this to a decimal by multiplying 0.125 by 100.

12.5%

They saved 12.5% by using the coupon.

1:10 1/10 10minutes/1meter HAH GET RATIOD BOZOD HAH

Can you f**king show a graph

multiply the 44 by 5/7 then divide by 2/5. EZ

I dont think such values exist. 

We can simplify the equations to:

\(4y+b = 0 \)

\(7x+ay=y-8\)

For both equations to have infinite solutions, the 2 equations must be multiples of each other. 

However, notice that the first equation has the slope \(-7 \over a\), and the second term has a slope of 0. 

Because the 2 slopes are different, there is no ordered pair (a,b) that satisfies the requirements. 

Wait.. I got a different answer. 

To factor the equation \(3x^2 + 14x + 8\), we need to find a pair of numbers that sum to 14 (middle number, coefficient of x) and multiply to 24 (product of 3 and 8, x^2 coefficient)

The 2 numbers that satisfy this pair are 2 and 12. Now, we rewrite the equation as: \(3x^2 + 12x +2x+ 8\)

Now, we can factor the first 2 terms and last 2 terms seperatelty. This gives us: \(3x(x+4) + 2(x+4)\). 

However, because both have an \((x+4)\) term in them, we can add them, giving us:\((3x+2)(x+4)\).

This means that \(A = 2\) and \(B = 4\), meaning \(A - B = 2 - 4 = \color{brown}\boxed{-2}\)

\(x^2 + y^2 = x^2 + (15 - 3x)^2 = 10x^2 - 90x + 225\)

Now you can complete the square and find the minimum. Can you take it from here?

Case 1: x²-5x+2 = 0

The roots are both real. The sum of the roots is equal to -b/a, which is 5 here.

Case 2: x²-4x+2=1

x²-4x+1=0

The roots are both real. The sum of the roots is equal to -b/a, which is 4 here.

Case 3: x²-4x+2=-1

x²-4x+3=0

The roots are both real. The sum of the roots is equal to -b/a, which is 4 here.

5+4+4 = 13

You are trying to turn it into a perfect square.

x²+5x+4=3x+m

x²+2x+(4-m)=0

This is a perfects square if m = 3. 

You're welcome.

Thank you. I can't believe I didn't see that before. 

No, please don't say that. Sometimes we miss the most obvious things. I think you will not miss it next time, so that is progress!

Let \((x, y) = \left(x, \dfrac1{\sqrt2}(x^2 - 8)\right)\) be a point on the graph.

Then the distance between the origin to that point is \(\sqrt{x^2 + \left(\dfrac1{\sqrt 2}(x^2- 8)\right)^2} = \sqrt{x^2 + \dfrac{(x^2 - 8)^2}2} = \sqrt{\dfrac{x^4 - 14x^2 + 64}2}\)

To minimize the distance, we need to minimize \(x^4 - 14x^2 + 64\). Note that \(x^4 - 14x^2 + 64 = (x^2 - 7)^2 + 15 \geq 15\), with equality attained when x = sqrt(7) or x = -sqrt(7). Then the minimum distance is \(\sqrt{\dfrac{15}2} = \dfrac{\sqrt{30}}2\).

Lol I'm the dumbest human of all time

I think there are 7 possible triangles:

1= (7, 7, 1)

2= (7, 6, 2)

3= (7, 5, 3)

4= (6, 6, 3)

5= (7, 4, 4)

6= (5, 6, 4)

7= (5, 5, 5)

Note that by the Pythagorean theorem, 

\(AC = \sqrt{10^2 + 20^2} = 10\sqrt5\)

Now, consider the interior angle sum of \(\triangle ABM\). We have \(\angle ABM + 90^\circ + \angle BAC = 180^\circ\). That means \(\angle ABM = 90^\circ - \angle BAC\)

Therefore, 

\(\begin{array}{rcl} \cos \angle ABM &=& \sin \angle BAC\\ &=& \dfrac{\boxed{\phantom{\text{aa}}}}{\boxed{\phantom{\text{aa}}}}\\ &=& \boxed{\phantom{\text{aaaa}}} \end{array}\)

Fill in the blanks. You should be able to arrive at the conclusion with the help of SOHCAHTOA.

Two sides are 11 and what? You need to include that number also.

Note that \(\operatorname{lcm}(a, b) = \dfrac{ab}{\gcd(a, b)}\), so if we want the smallest value of lcm, we want to minimize the product a * b.

That means a and b must be as small as possible.

The smallest number with unit digit 2 and divisible by 8 is 32.

The smallest number with unit digit 4 and divisible by 8 is 24.

Can you continue from here?