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2 years ago

What is 1+1? Very urgent

Mathematics

1 answer:

forsale [732]2 years ago

8 0

Answer:

Step-by-step explanation:

anything plus 1 equals 62.

Hope this helps. Have a nice day you amazing bean child.

( jk jk. It's 2)

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Solve for m 2/5 = m/70 = show work and order of operations

Black_prince [1.1K]

Answer:

m = 28

Step-by-step explanation:

70 × 2 = 5m

140 = 5m

28 = m

4 0

3 years ago

Khan Academy Volume of Sphere Answers!!!

agasfer [191]

Explanation:

The N stands for Pi so when you type in the answers to Khan Academy use the Pi symbol when theres an N! These are the common questions and common radius Khan Academy gives when they ask to find the radius of the circle.

Answers:

For 4 - 267.95

For 1/4 - 1/48n

For 1 - 4/3n

For 9 - 972n

For 7 - 1372/3n

For 6 - 288n

For 3 - 36n

For 10 - 4000/3n

For 8 - 2048/3n

For 1/2 - 1/6n

For 5 - 500/3n

For 2 - 33.49

Step-by-step explanation:

I did the go math so all of these rights and the majority of the answers are the questions Khan Academy will ask you to find the radius of the circle.

5 0

1 year ago

A plumber has a a 36-in. pipe. He must cut it into two pieces so that one piece is 14 inches longer that the other. How long is

irakobra [83]

You have to do this 32-14=12

One piece is 14” and the second one is 12”

4 0

3 years ago

Here is quite a simple question. How do you work out the area of a circle?

Tomtit [17]

$A=\pi r^2$ where r is the radius of a circle.

7 0

2 years ago

Read 2 more answers

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.

8 0

3 years ago