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3 years ago

What is the answers to this math worksheet ?

Mathematics

1 answer:

tester [92]3 years ago

7 0

For #1 the median is 16

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How to add and subtract polynomials

Lorico [155]

You combine like terms

Example: 5x^3+3x+1-4x^3-2x-1

The answer would be x^3+x

6 0

3 years ago

Roberto and Maria Santanos spend approximately $10.75 per week to wash and dry their family's clothes at a local coin laundry. A

timama [110]

Answer:

72 weeks

Step-by-step explanation:

If 1 week =$10.75 then $774 will be equal to $774×1 week=774 then divide by $10.75=72 weeks

4 0

4 years ago

Please help me with this question ASP

Wewaii [24]

Answer:

5/6 pounds

Step-by-step explanation:

2 1/2=5/2. 1/3 of 5/2=1/3•5/2. 1/3•5/2=5/6.

3 0

3 years ago

Read 2 more answers

when is it possible for the product of a rational and an irrational number to result in a rational value?

zimovet [89]

It is always possible for the product of a rational number and irrational number to result in a rational value.

7 0

4 years ago

P(k)=a^k=2 3 4 find value of a that makes this is a valid probability distribution

Vesna [10]

Sounds like you're asked to find $a$ such that

$\displaystyle\sum_{k=2}^4\mathbb P(k)=\mathbb P(2)+\mathbb P(3)+\mathbb P(4)=1$

In other words, find $a$ that satisfies

$a^2+a^3+a^4=1$

We can factorize this as

$a^4+a^3+a^2-1=a^3(a+1)+(a-1)(a+1)=(a+1)(a^3+a-1)=0$

In order that $\mathbb P(k)$ describes a probability distribution, require that $\mathbb P(k)\ge0$ for all $k$ , which means we can ignore the possibility of $a=-1$ .

Let $a=y+\dfrac xy$ .

$a^3+a-1=\left(y+\dfrac xy\right)^3+\left(y+\dfrac xy\right)-1=0$
$\left(y^3+3xy+\dfrac{3x^2}y+\dfrac{x^3}{y^3}\right)+\left(y+\dfrac xy\right)-1=0$

Multiply both sides by $y^3$ .

$y^6+3xy^4+3x^2y^2+x^3+y^4+xy^2-y^3=0$

We want to find $x\neq0$ that removes the quartic and quadratic terms from the equation, i.e.

$\begin{cases}3x+1=0\\3x^2+x=0\end{cases}\implies x=-\dfrac13$

so the cubic above transforms to

$y^6-y^3-\dfrac1{27}=0$

Substitute $y^3=z$ and we get

$z^2-z-\dfrac1{27}=0\implies z=\dfrac{9+\sqrt{93}}{18}$
$\implies y=\sqrt[3]{\dfrac{9+\sqrt{93}}{18}}$
$\implies a=\sqrt[3]{\dfrac{9+\sqrt{93}}{18}}-\dfrac13\sqrt[3]{\dfrac{18}{9+\sqrt{93}}}$

6 0

4 years ago