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In simplest radical form, what are the solutions to the quadratic equation 6 = x2 – 10x?

FrozenT [24]

Quadratic equation: ax² + bx + c =0

x' = [-b+√(b²-4ac)]/2a and x" = [-b-√(b²-4ac)]/2a

6 = x² – 10x ; x² - 10x -6 =0
(a=1, b= - 10 and c = - 6

x' = [10+√(10²+4(1)(-6)]/2(1) and x" = [10-√(10²+4(1)(-6)]/2(1)
x' =5+√31 and x' = 5-√31

7 0

1 year ago

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The junior class has 50 students. Twenty-five students take only french, 15 take only Latin, and 10 take both. If a student is c

zavuch27 [327]

There are 15 students who take Latin, and 10 students who take both.
Therefore:
15+10 =25 total students taking Latin
25students / 50 total students = 1/2

The answer is 1/2.

5 0

2 years ago

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se the function to show that fx(0, 0) and fy(0, 0) both exist, but that f is not differentiable at (0, 0). f(x, y) = 9x2y x4 + y

alexandr1967 [171]

Answer:

It is proved that $f_x, f_y$ exixts at (0,0) but not differentiable there.

Step-by-step explanation:

Given function is,

$f(x,y)=\frac{9x^2y}{x^4+y^2}; (x,y)\neq (0,0)$

To show exixtance of $f_x(0,0), f_y(0,0)$ we take,

$f_x(0,0)=\lim_{h\to 0}\frac{f(h+0,k+0)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{9h^2k}{h^4+k^2}-0}{h}\\\therefore f_x(0,0)=\lim_{h\to 0}\frac{9hk}{h^4+k^2}=\lim_{h\to 0}\frac{9k}{h^3+\frac{k^2}{h}}=0$ exists.

And,

$f_y(0,0)=\lim_{k\to 0}\frac{f(h,k)-f(0,0)}{k}=\lim_{k\to 0}\frac{9h^2k}{k(h^4+k^2)}=\lim_{k\to 0}\frac{9h^2}{h^4+k^2}=\frac{9}{h^2}$ exists.

To show f(x,y) is not differentiable at the origin cheaking continuity at origin be such that,

$\lim_{(x,y)\to (0,0)}\frac{9x^2y}{x^4+y^2}=\lim_{x\to 0\\ y=mx^2}\frac{9x^2y}{x^4+y^2}=\frac{9x^2\times m x^2}{x^4+m^2x^4}=\frac{9m}{1+m^2}$ where m is a variable.

which depends on various values of m, therefore limit does not exists. So f(x,y) is not continuous at (0,0). Hence it is not differentiable at (0,0).

4 0

2 years ago

Match the following items by evaluating the expression for x = -6.

Kryger [21]

Step-by-step explanation:

Put the value of x = -6 to all expressions:

$x^{-2}=(-6)^{-2}=\dfrac{1}{(-6)^2}=\dfrac{1}{6}\qquad\text{used}\ a^{-n}=\dfrac{1}{a^n}\\=====================\\x^{-1}=(-6)^{-1}=\dfrac{1}{(-6)^1}=-\dfrac{1}{6}\\=====================\\x^0=(-6)^0=1\qquad a^0=1\ \text{for all real numbers except 0}\\=====================\\x^1=(-6)^1=-6\qquad a^1=a\ \text{for all real numbers}\\=====================\\x^2=(-6)^2=36$

5 0

2 years ago

Consider the angle shown below with an initial ray pointing in the 3-o'clock direction that measures θ radians (where 0≤θ<2π)

neonofarm [45]

Answer:

hello attached below is the sketch of the missing diagram

a) h = 2

b) cos^-1 ( 2 ) = ∞

c ) θ ≈ 215°

Step-by-step explanation:

a) The length ( h ) to the right of the circles center

= h^2 = ( -1.15)^2 + ( -1.64)^2

∴ h = 2

b) cos^-1 ( 2 ) = ∞

c) The value gotten from b does not give us the value of θ

AB = - 1.64

BC = -1.15

AC = 2

sin^-1 ( - 1.64 / -1.15 )

cos^1 ( -1.15 / 2 ) gives us the value of θ

tan^-1 ( -1.64 / -1.15 )

∴ θ ≈ 215°

6 0

2 years ago