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3 years ago

ME LO RESUELVEN PORFAA

Mathematics

1 answer:

EastWind [94]3 years ago

3 0

The hypotenuseb of the triangle is 17.

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Here is an expression: (7x + ¼)3

julia-pushkina [17]

Answer: 7x + 13/4

Step-by-step explanation:

There you go

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3 years ago

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Which statement best describes the two figures

ollegr [7]

The answer would most likely be reflection

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3 years ago

What percent of 75 is 15?

lora16 [44]

20% of 75 is 15
hope this helps

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3 years ago

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Factor 9x^+12x+4=0 please help

UkoKoshka [18]

Answer:

(3x+2) ^2

x = -2/3

Step-by-step explanation:

9x^2+12x+4 = 0

We need to recognize that this is perfect squares

(a + b)2 = a^2 + 2ab + b^2;

Where a = 3x and b =2

a^2 = 9x^2 2ab = 2*3x*2 = 12x and b^2 = 2^2 =4

So we can factor this as (3x+2) ^2

If you want to solve

We can use the zero product property

(3x+2) ^2 =0

Take the square root of each side

3x+2 =0

Subtract 2 from each side

3x=-2

Divide by 3

3x/3 = -2/3

5 0

3 years ago

The equations that must be solved for maximum or minimum values of a differentiable function w=f(x,y,z) subject to two constrai

sammy [17]

The Lagrangian is

$L(x,y,z,\lambda,\mu)=x^2+y^2+z^2+\lambda(4x^2+4y^2-z^2)+\mu(2x+4z-2)$

with partial derivatives (set equal to 0)

$L_x=2x+8\lambda x+2\mu=0\implies x(1+4\lambda)+\mu=0$

$L_y=2y+8\lambda y=0\implies y(1+4\lambda)=0$

$L_z=2z-2\lambda z+4\mu=0\implies z(1-\lambda)+2\mu=0$

$L_\lambda=4x^2+4y^2-z^2=0$

$L_\mu=2x+4z-2=0\implies x+2z=1$

Case 1: If $y=0$ , then

$4x^2-z^2=0\implies4x^2=z^2\implies2|x|=|z|$

Then

$x+2z=1\implies x=1-2z\implies2|1-2z|=|z|\implies z=\dfrac25\text{ or }z=\dfrac23$

$\implies x=\dfrac15\text{ or }x=-\dfrac13$

So we have two critical points, $\left(\dfrac15,0,\dfrac25\right)$ and $\left(-\dfrac13,0,\dfrac23\right)$

Case 2: If $\lambda=-\dfrac14$ , then in the first equation we get

$x(1+4\lambda)+\mu=\mu=0$

and from the third equation,

$z(1-\lambda)+2\mu=\dfrac54z=0\implies z=0$

Then

$x+2z=1\implies x=1$

$4x^2+4y^2-z^2=0\implies1+y^2=0$

but there are no real solutions for $y$ , so this case yields no additional critical points.

So at the two critical points we've found, we get extreme values of

$f\left(\dfrac15,0,\dfrac25\right)=\dfrac15$ (min)

and

$f\left(-\dfrac13,0,\dfrac23\right)=\dfrac59$ (max)

5 0

4 years ago