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3 years ago

One foot is equal to 12 inches. The function I(f) takes a length in feet (as input) and returns a length in inches (as output).

1 answer:

8 0

You need to substitute the number 6 into the function. The output would be 72, because when l(6) = 12(6), you would get l(6) = 72.

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Activity calculate the mean of each set of numbers. a) 54,35, 27, 61, 38

STALIN [3.7K]

Answer:

The Mean is 43

Step-by-step explanation:

If you need me to explain how I got there- I would be happy to ^w^

4 0

2 years ago

Read 2 more answers

Consider the function f(x, y) = x2 + xy + y2 defined on the unit disc, namely, d = {(x, y)| x2 + y2 ≤ 1}. use the method of lagr

Pani-rosa [81]

First we note that the partial derivatives vanish simultaneously at one point:

$\begin{cases}f_x=2x+y=0\\f_y=x+2y=0\end{cases}\implies(x,y)=(0,0)$

so there is one critical point within the region $D$ .

The Lagrangian is

$L(x,y,\lambda)=x^2+xy+y^2+\lambda(x^2+y^2-1)$

and has partial derivatives

$L_x=2x+y+2\lambda x$
$L_y=x+2y+2\lambda y$
$L_\lambda=x^2+y^2-1$

Set each partial derivative to 0 to find the possible critical points within the disk $D$ . Then we notice that

$yL_x=2xy+y^2+2\lambda xy=0$
$xL_y=x^2+2xy+2\lambda xy=0$
$L_\lambda=0\implies x^2+y^2=1$

$\implies xL_y-yL_x=x^2-y^2=0$

Since $x^2+y^2=1$ , we have

$x^2-y^2=x^2+y^2-2y^2=0\implies1=2y^2\implies y^2=\dfrac12\implies y=\pm\dfrac1{\sqrt2}$

And since $x^2-y^2=0$ , or $x^2=y^2$ , we also have

$x=\pm\dfrac1{\sqrt2}$

So we have four possible additional critical points to consider:

$f(0,0)=0$
$f\left(-\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)=\dfrac32$
$f\left(-\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)=\dfrac12$
$f\left(\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)=\dfrac12$
$f\left(\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)=\dfrac32$

It should be clear enough which of these correspond to the absolute extrema of $f$ over $D$ .

8 0

3 years ago

Which shape has four congruent sides and two sets of parallel sides?

Dovator [93]

It’s a Parallelogram

6 0

2 years ago

HELP ME ITS TIMED AND I WILL GIVE BRAINLIEST!!!!!

Olegator [25]

To find the gradient of a line we first hv to find the formula

so Gradient=√X1-x2and √y1-y2

((-2,1) and (0,-5)) line 1

line 2(0,-1) and (1,0)

first we will work for line 1

X1 is -2

X2 is -0

Y1 is 1

Y2 is -1

so y=x-1

so we substitute the values

1--1=-2--0

so,

1+1=-2+0

2=-2

-0

So the equation 1 line 1 is not true

we will now go to equation 2 line 2

the equation is 3x+y=-5

we substitute the values

3(-2--0)+1--1=-5

3(-2)+-2=-5

-6+-2=-5

-8= -5

group like terms

Ur answer should look like this -3

So the answer is 3x+y=-5

7 0

3 years ago

I need to know the answer please , thank you !

n200080 [17]

The answer is D your welcome

3 0

3 years ago