(a^2 - b^2)=???????????????????????????

Which graph best represents the relationship between time and the number of teams registered?

alexdok [17]

<h2>Explanation:</h2><h2></h2>

The complete question is in the attached file. So we have to choose between two graphs. On of them is a linear model while the other is an exponential model. From the statements, we have a relationship between time and the number of teams registered. So we can establishes variables in the following form:

$x:\text{Time} \\ \\ y:\text{Number of teams registered}$

We also know that each week 6 teams register to participate, so:

$\bullet \ \text{For week 0:} \rightarrow \text{0 teams registered} \\ \\ \bullet \ \text{For week 1:} \rightarrow \text{6 teams registered} \\ \\ \bullet \ \text{For week 2:} \rightarrow \text{12 teams registered (Because 6+12)} \\ \\ \bullet \ \text{For week 3:} \rightarrow \text{18 teams registered (Because 12+6)}$

As you can see, as x increases one week, y increases at a constant ratio of 6. Therefore, this can be modeled by a linear function given by the form:

$y=6x$

In conclusion, <em>the linear model (first graph below) is the one that bests represents the relationship between time and the number of teams registered.</em>

7 0

3 years ago

Given that an intercepted arc has length 6pi inches and a central angle is pi/3, find the radius

MA_775_DIABLO [31]

$\bf \textit{arc's length}\\\\
s=r\theta \quad 
\begin{cases}
r=radius\\
\theta =angle~in\\
\qquad radians\\
------\\
s=6\pi \\
\theta =\frac{\pi }{3}
\end{cases}\implies 6\pi =r\left( \frac{\pi }{3} \right)\implies \cfrac{6\pi }{\frac{\pi }{3}}=r
\\\\\\
\cfrac{6\pi }{1}\cdot \cfrac{3}{\pi }=r\implies 18=r$

5 0

3 years ago

What is the unit rate and state rate of 253 for 20 hours?

Vilka [71]

The unit rate is 12.65 per hour.

6 0

3 years ago

1. Alicia has 7 paintings she wants to hang side by side on her wall.

Kipish [7]

A.

If we take 7 paintings to be hung in 7 spaces side by side, the first space can have any one of the 7 paintings, the second space can have any one of the remaining 6 paintings (as 1 is already hung), the third space can have any one of the remaining 5 paintings (as 2 already hung)...It goes on like this.

So we have $7*6*5*4*3*2*1=5040$ ways to arrange all the paintings from left to write. <em>(in factorial notation it is 7!=5040)</em>

B.

We use combinations rather than permutations because order doesn't matter. If we name the paintings A,B,C,D,E,F, and G, groups of 3 paintings of ABC or ACB are the same. So we evaluate $7C3$ using the combination formula,

$nCr=\frac{n!}{(n-r)!r!}$

We have,

$7C3=\frac{7!}{(7-3)!*3!} = \frac{7!}{4!*3!} = 35$

C.

This is similar to part A in some ways. Any 3 pictures can be arranged in $3!$ different ways. $3!=3*2*1=6$ . So, 6 different ways.

ANSWER:

A) 5040 ways

B) 35 different groups

C) 6 ways

8 0

3 years ago

Read 2 more answers

X^6-9x^4-x^2+9=0 please help me I don't understand this

Rom4ik [11]

<h2>Steps:</h2>

So firstly, I will be factoring by grouping. For this, factor x⁶ - 9x⁴ and -x² + 9 separately. Make sure that they have the same quantity on the inside of the parentheses:

$x^4(x^2-9)-1(x^2-9)=0$