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3 years ago

Yes i give brainliest

Mathematics

1 answer:

satela [25.4K]3 years ago

8 0

D

It can give some balls more chance to be picked, making it unfair

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Help please? What are the factors of the product represented below?

Ilia_Sergeevich [38]

Answer:

Step-by-step explanation:

count how many x^2 there are because they would tell you what your answer is.

8 0

3 years ago

Read 2 more answers

Solve the system of equations. 13x−y=90

Ann [662]

Lets solve the system of equations step by step.
$13x-y=90$ equation (1)
$y=x^2-x-42$ equation (2)

Step 1. Solve for $y$ in equation (1)
$13x-y=90$
$-y=90-13x$
$y=13x-90$ equation (3)

Step 2. Replace equation (3) in equation (2) and solve for $x$ :
$y=x^2-x-42$
$13x-90=x^2-x-42$
$x^2-14x+48=0$
$(x-6)(x-8)=0$
$x=6$ equation (4)
$x=8$ equation (5)

Step 3. Replace equation (4) and equation (5) in equation (3)
(4) in (3)
$y=13x-90$
$y=13(6)-90$
$y=-12$

(5) in (3)
$y=13x-90$
$y=13(8)-90$
$y=14$

We can conclude that the solutions of our system of equations are (6,-12) and (8,14)

8 0

4 years ago

Read 2 more answers

X^y=y^x find derivitive

Igoryamba

Answer:

$y'=\frac{y^2-xy\ln(y)}{x^2-xy\ln(x)}$

Step-by-step explanation:

Take natural log of both sides first.

$x^y=y^x$

$\ln(x^y)=\ln(y^x)$

Taking the natural log of both sides allows you to bring down the powers.

$y\ln(x)=x\ln(y)$

I'm going to differentiate both sides using the power rule.

$(y)'(\ln(x))+(\ln(x))'y=(x)'(\ln(y))+(\ln(y))'x$

Now recall (ln(x))'=(x)'/x=1/x while (ln(y))'=(y)'/y=y'/y.

$y'(\ln(x))+\frac{1}{x}y=1(\ln(y))+\frac{y'}{y}x$

Simplifying a bit:

$y' \ln(x)+\frac{y}{x}=\ln(y)+\frac{y'}{y}x$

Now going to gather my terms with y' on one side while gathering other terms without y' on the opposing side.

Subtracting y'ln(x) and ln(y) on both sides gives:

$\frac{y}{x}-\ln(y)=-y'\ln(x)+\frac{y'}{y}x$

Now I'm going to factor out the y' on the right hand side:

$\frac{y}{x}-\ln(y)=(-\ln(x)+\frac{x}{y})y'$

Now we get to get y' by itself by dividing both sides by (-ln(x)+x/y):

$\frac{\frac{y}{x}-\ln(y)}{-\ln(x)+\frac{x}{y}}=y'$

Now this looks nasty to write mini-fractions inside a bigger fraction.

So we are going to multiply top and bottom by xy giving us:

$\frac{y^2-yx\ln(y)}{-xy\ln(x)+x^2}=y'$

$y'=\frac{y^2-xy\ln(y)}{x^2-xy\ln(x)}$

5 0

3 years ago

What is a number that when you subtract 12 and then divide the difference by 3 will equal 6?

Viktor [21]

Answer:

105 is ur answer

Step-by-step explanation:

3 0

2 years ago

(40 POINTS!)

Ainat [17]

Hello there!

The center of any circle is the midpoint of the diameter.
Let's use the midpoint formula to find the midpoint of the diameter.

The midpoint of a line segments with endpoints $(x_1, y_2)$ and $(x_2, y_2)$ will have a midpoint of $( \dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2})$

Midpoint of the diameter:

$( \dfrac{-3-11}{2}, \dfrac{-5-7}{2})$

$=(-7, -6)$

Your answer is B. Hope this helps!

3 0

3 years ago

Yes i give brainliest​

Yes i give brainliest