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What is the coefficient of x^3 in the expansion of (x + 2)^7?

Hallo Gast!

\((x+2)^7=x^7+14x^6+...+{\color{blue}560x^3}+...+128\)

  !

This is very simple...

[LPQ] = √20(3√20)

[LPQ] / [JKL] = 3/16

[PQR] / [JKL] = 1/16

CD = sqrt(BC² - BD²)

sin(B) = CD / BC

Let ABC be the right-angled isosceles triangle with angle A = 90 degrees.

AM = BM = √50

MD = 1/2(AM)

Angle AME = 45º

Tangent of angle BDM = 2

Find TD and TM, and then using Heron's formula, calculate the area of triangle DME. 

MN is NOT equal to 10 !!! 

MN = 14 / 2 = 7 !!! 

Find u/v + v/u.

Hello Guest!

\(3x^2 + 5x + 7 = 2x^2 - 8x + 1\\ x^2+13x+6=0\)

\(x=-6.5\pm\sqrt{6.5^2-6}\\ x=-6.5\pm \sqrt{36.25}\\ \color{blue}u=-6.5+ \sqrt{36.25}\\ \color{blue}v= -6.5- \sqrt{36.25}\)

\({\color{blue}\frac{u}{v}}=\frac{-6.5+ \sqrt{36.25}}{-6.5- \sqrt{36.25}}\color{blue}=0.0382725\\ {\color{blue}\frac{v}{u}}=\frac{-6.5- \sqrt{36.25}}{-6.5+\sqrt{36.25}}\color{blue}=26.12839\)

  !

Radii are 7, 7, and 11 

4.3           pH = - log [H+]  

                         - log ( 5.22 x 10^-5) = 4.3

3^5/27^3

3^5  / ( 3^3)^3

3^5 / 3^9

3^-4   for the right side

for the left side

(3^2)^y 

so    3^(2y) = 3^(-4)      can you finish from here (Hint: equate the exponents)

Can't see the figure....

Find a monic quartic polynomial with rational coefficients whose roots include

x = 1 - sqrt(2) and x = 2 + sqrt(7).

Hello Guest!

\(f(x)=(x-1+\sqrt{2})(x-2-\sqrt{7})\\ f(x)= x^2-\sqrt{7}\ x+\sqrt{2}\ x-3x-\sqrt{14}+\sqrt{7}-2\sqrt{2}+2\)

\(f(x)=x^2-(\sqrt{7}-\sqrt{2}+3)x-\sqrt{14}+\sqrt{7}-2\sqrt{2}+2=0\)

\(g(x)=x^2-4.2315x-1.9243\approx 0\)

  !

this is x = 9.8

Hello Guest,

\(\frac{5}{4}k-\frac{3}{4}+\frac{1}{2}k=-\frac{1}{4}k-\frac{1}{2}\)

\(\mbox {Now multiply both sides by 4}\)

\(5k-3+2k=-k-2\)

\(7k-3=-k-2\)

\(7k-3+k=-2\)

\(7k+k=-2+3\)

\(8k=-2+3\)

\(8k=1\)

\(k=\frac{1}{8}=0.125\)

Easy!

Hello Guest,

\((x,y)=(0,0)\) AND

\((x,y)=(245,49)\)

Hello Guest,

so that's an easy question:

\(3.5x-5.7=2.5x+4.1\)

\(3.5x-5.7-2.5x=4.1\)

\(3.5x-2.5x=4.1+5.7\)

\(x=4.1+5.7\)

\(x=9.8=\frac{49}{5}=9\frac{4}{5}\)

Easy!

okay, thanks!

You are correct! NX = 5, and MN is a midsegment of triangle ABX.

MN = 1/2(BX) 

AX = 10 

MN = 10.

365 days/yr

70 oz/day   *  365 day/yr   = 70 * 365 =  ..........   oz/yr

(x -1-sqrt2)(x-1+sqrt2)(x-2-sqrt7)(x-2+sqrt7) =

x⁴ -6x³ + 4x² +10x + 3

Suppose that the common roots are a and b and that the other, individual roots, are m and n.

We then have (Veita),

a + b + m = -5, .................(1)

a + b + n  =  5, .................(2)

ab + bm + ma = p, ...........(3)

ab + bn + na = p, .............(4).

So, 

(2) - (1):  n - m = 10, ........(5)

(4) - (3):  b(n - m) + a(n - m) = 0, .....(6).

from which

b + a = 0

and then, from (1) and (2),

m = -5 and n = 5.

\(\)

y = mx + b   

We can see that the slope is 2 and the y-intercept is 3, so  

y = 2x + 3  

2x – y = –3 

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