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3 years ago

What is the probability of picking a purple marble and rolling an odd number?

Mathematics

2 answers:

Lostsunrise [7]3 years ago

6 0

5/11 for the marble and ½ for dice

Vaselesa [24]3 years ago

4 0

Answer:

5/11

Step-by-step explanation:

I hope this helps a bit.

You might be interested in

Find the value of k such that x-2 is a factor of 3x³-kx²+5x+k

Nuetrik [128]

Answer:

$\displaystyle k = \frac{34}{3}$

Step-by-step explanation:

We are given the polynomial:

$\displaystyle P(x) = 3x^3 - kx^2 + 5x + k$

And we want to determine the value of k such that (x - 2) is a factor of the polynomial.

Recall that the Factor Theorem states that a binomial (x - a) is a factor of a polynomial P(x) if and only if P(a) = 0.

Our binomial factor is (x - 2). Thus, a = 2.

Hence, by the Factor Theorem, P(2) must equal zero.

Find P(2):

$\displaystyle \begin{aligned} P(2) &= 3(2)^3 - k(2)^2 + 5(2) + k \\ \\ &= 3(8) - 4k + 10 + k \\ \\ &= 34 - 3k \end{aligned}$

This must equal zero. Hence:

$\displaystyle \begin{aligned} 34 - 3k &= 0 \\ \\ -3k &= -34 \\ \\ k = \frac{34}{3} \end{aligned}$

In conclusion, k = 34/3.

6 0

2 years ago

URGENT I WILL MARK BRAINLIEST DRAW WHERE THE DOTS GO ON THE GRAPH

frutty [35]

Answer:

(Answer in picture)

3 0

2 years ago

Write the equation -4x^2+9y^2+32x+36y-64=0 in standard form. Please show me each step of the process!

IgorC [24]

Hey there, hope I can help!

$-4x^2+9y^2+32x+36y-64=0$

$\mathrm{Add\:}64\mathrm{\:to\:both\:sides} \ \textgreater \ 9y^2+32x+36y-4x^2=64$

$\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \ -4\left(x^2-8x\right)+9\left(y^2+4y\right)=64$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4$
$-\left(x^2-8x\right)+\frac{9}{4}\left(y^2+4y\right)=16$

$\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9$
$-\frac{1}{9}\left(x^2-8x\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}$

$\mathrm{Convert}\:x\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x^2-8x+16\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)$

$\mathrm{Convert}\:y\:\mathrm{to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y+4\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Convert\:to\:square\:form}$
$-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)$

$\mathrm{Refine\:}\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right) \ \textgreater \ -\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=1$

$Refine\;once\;more\;-\frac{\left(x-4\right)^2}{9}+\frac{\left(y+2\right)^2}{4}=1$

For me I used
$\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}= 1$
$As\;\mathrm{it\;\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}$

I know yours is an equation which is why I did not go any further because this is the standard form you are looking for. I would rewrite mine to get my hyperbola standard form. However the one I have provided is the form you need where mine would be.
$\frac{\left(y-\left(-2\right)\right)^2}{2^2}-\frac{\left(x-4\right)^2}{3^2}=1$

Hope this helps!

4 0

3 years ago

Need help plz very difficult

Oksanka [162]

Answer:

D - 9x^(2)+2x-5

Step-by-step explanation:

6 0

3 years ago

What type of polynomial is this

Crank

Answer:

trinomial..............

5 0

3 years ago