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3 years ago

Evaluate the following expression (a + b)^0

Mathematics

2 answers:

ira [324]3 years ago

8 0

Answer:

the answer is 1

elixir [45]3 years ago

3 0

Answer:

So, $(a + b)^0=1$

Step-by-step explanation:

$(a + b)^0$

To evaluete this expression we apply exponential property

$x^0=1$

IF any expression has exponent 0 then the value is 1 always

for example, $3^0 =1$

$5^0=1$

$(x+2)^0=1$ because exponent is 0

So, $(a + b)^0=1$

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(-w-6)/(w^2+4w-12), how do i write this into simplest form?

PIT_PIT [208]

First factorise the top = -(w+6)
The first factorise the bottom: (w^2 + 4w -12) = (w+6) (w-2)
Therefore the two (w+6) cancels out and you are left with -1 / w-2

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3 years ago

Need help

ira [324]

Formula for the volume of a cylinder is pi radius squared times height so this one would be pi times 25 times 4 which approximately equals 314

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HELP PLEASE 18 miles= how many feet _________feet

Delvig [45]

Answer:

18 miles= how many feet

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Step-by-step explanation:

happy to help ya:)

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3 years ago

If a person is 5 1/2 tall and cast a 3 1/2 shadow And there is a flag poll that is 28 feet tall what is the height of the shadow

Serjik [45]

Answer: 17 9/11

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3 years ago

Find the minimum and maximum values of the function subject to the given constraint. (if an answer does not exist, enter dne.) f

nata0808 [166]

Via Lagrange multipliers:

$L(x,y,\lambd)=x^2y+x+y+\lambda(xy-5)$
$L_x=2xy+1+\lambda y=0$
$L_y=x^2+1+\lambda x=0$
$L_\lambda=xy-5=0$

$\underbrace{10}_{2xy}+1+\lambda y=0\implies \lambda=-\dfrac{11}y$
$xy=5\implies y=\dfrac5x\implies\lambda=-\dfrac{11}5x$

$\impliesx^2+1+\left(-\dfrac{11}5x\right)x=0\implies x^2=\dfrac56\implies x=\pm\sqrt{\dfrac56}$
$xy=5\implies y=\pm\sqrt{30}$

At these points, we get local minima of $f\left(\pm\sqrt{\dfrac56},\pm\sqrt{30}\right)=\pm2\sqrt{30}$ .

- - -

Another way to do this is to make $f(x,y)$ a function independent of $y$ , which is made possible by the constraint.

$xy=5\implies y=\dfrac5x$
$\implies f(x,y)=f\left(x,\dfrac5x\right)=F(x)=6x+\dfrac5x$
$\implies F'(x)=6-\dfrac5{x^2}=0\implies x=\pm\sqrt{\dfrac56}$

and so on.

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3 years ago