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3 years ago

What is jk? please help

Mathematics

1 answer:

motikmotik3 years ago

3 0

Answer:

it means just kidding

Step-by-step explanation:

pls mark brainliest

You might be interested in

Which order pair is a solution to the system of linear equations -4+y=8 and x-5y=17?

andreev551 [17]

Hello!

To find this ordered pair, we solve for x and y and put them together in an ordered pair.

-4+y=8

x-5y=17

Let's solve the first equation for y.

y= 12

Now, let's plug 12 into the second equation for y.

x-5(12)=17

x-60=17

x=77

Therefore, our ordered pair is (77,12)

I hope this helps!

4 0

3 years ago

Read 2 more answers

PLEASE HELP ME!!

Step2247 [10]

Answer:

A = $10,441.68

A = P + I where

P (principal) = $10,400.00

I (interest) = $41.68

Step-by-step explanation:

First, convert R as a percent to r as a decimal

r = R/100

r = 0.02/100

r = 0.0002 rate per year,

Then solve the equation for A

A = P(1 + r/n)nt

A = 10,400.00(1 + 0.0002/2)(2)(20)

A = 10,400.00(1 + 0.0001)(40)

A = $10,441.68

Summary:

The total amount accrued, principal plus interest, with compound interest on a principal of $10,400.00 at a rate of 0.02% per year compounded 2 times per year over 20 years is $10,441.68.

8 0

3 years ago

Find the value of x in 2x+20=3x-8

Savatey [412]

Answer:

X=28

Step-by-step explanation:

3x-8=2x+20. So x=28

8 0

3 years ago

Read 2 more answers

Find the absolute maximum and absolute minimum values of the function f(x, y) = x 2 + y 2 − x 2 y + 7 on the set d = {(x, y) : |

dsp73

Looks like $f(x,y)=x^2+y^2-x^2y+7$ .

$f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1$

$f_y=2y-x^2=0\implies2y=x^2$

If $x=0$ , then $y=0$ - critical point at (0, 0).
If $y=1$ , then $x=\pm\sqrt2$ - two critical points at $(-\sqrt2,1)$ and $(\sqrt2,1)$

The latter two critical points occur outside of $D$ since $|\pm\sqrt2|>1$ so we ignore those points.

The Hessian matrix for this function is

$H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}$

The value of its determinant at (0, 0) is $\det H(0,0)=4>0$ , which means a minimum occurs at the point, and we have $f(0,0)=7$ .

Now consider each boundary:

If $x=1$ , then

$f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4$

which has 3 extreme values over the interval $-1\le y\le1$ of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).

If $x=-1$ , then

$f(-1,y)=8-y+y^2$

and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).

If $y=1$ , then

$f(x,1)=8$

which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).

If $y=-1$ , then

$f(x,-1)=2x^2+8$

which has 3 extreme values, but the previous cases already include them.

Hence $f(x,y)$ has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).

3 0

3 years ago

Hannah has another candle that is 14 cm tall. How fast must it burn in order to also be 6 cm tall after 4 hours? Explain your th

RideAnS [48]

Answer:

The candle must burn at 2cm per hour in order to be 6cm tall after 4 hours. I know this because the candle has to burn 8cm in 4 hours and that amounts to 2cm per hour.

Step-by-step explanation:

Lets solve!

We have to find out how much of the candle needs to be burned!

14cm - 6cm = 8cm

The candle must burn 8cm in 4 hours!

Lets find out how much the candle burns in 1 hour.

$\frac{8cm}{4hours}$ = 2cm per hour

The candle must burn at 2cm per hour in order to be 6cm tall after 4 hours. I know this because the candle has to burn 8cm in 4 hours and that amounts to 2cm per hour.

Woohoo, We did it! <u>Would you like to mark my answer as brainliest?</u> I would love that!

6 0

3 years ago