All categories

3 years ago

2 Points What can you say about the end behavior of the function f(x)=-4x + 6x2 - 52?

1 answer:

3 0

Just look at coefficient of highest power of x (leading coefficient): 6. It is x^2, so limf(x) x—>infinity and limf(x) x—>-infinity are both infinity

You might be interested in

A 2-ft vertical post casts a 20-in shadow at the same time a nearby cell phone tower casts a 125-ft shadow. How tall is the c

chubhunter [2.5K]

Answer:

1249. 5 inches

Step-by-step explanation:

Pole length , Pl = 2 fts

Pole shadow, Ps = 20 in = (0.0833 * 20) = 1.666 ft

Tower length, Tl = 125 ft

Tower shadow, Ts = x

Pl / Ps = Tl/ Ts

2/ 1.666 = 125 / x

Cross multiply

2x = 125 * 1.666

2x = 208.25

x = 104.125 ft

x = 104.125 * 12 = 1249. 5 inches

5 0

3 years ago

(a) Each student in a pottery class is responsible for cleaning the area around his or her workspace at the end of the lesson. E

ArbitrLikvidat [17]

Ok this is not as hard as it seems

(4.5 feet)^2 times the number of students (13)

20.25 x 13

263.25 is the correct answer

I hope it helps :<

5 0

3 years ago

Find the smallest number that is exactly divisible by 144 192 and 168

svlad2 [7]

Answer:

2 is your answer. hope helpful answer

3 0

3 years ago

Eliminate the parameter. x = t2 + 2, y = t2 - 4

stepan [7]

$\bf \begin{cases}
x=t^2+2\implies x-2=t^2\implies \sqrt{x-2}=\boxed{t}\\\\
y=t^2-4\\
----------\\
y=\left( \boxed{\sqrt{x-2}} \right)^2-4\implies y=x-2-4\\\\
y=x-6
\end{cases}$

just a plain vanilla substitution

8 0

3 years ago

Read 2 more answers

4. A random variable X has a mean of 10 and a standard deviation of 3. If 2 is added to each value of X, what will the new mean

Ede4ka [16]

Adding 2 to each value of the random variable $X$ makes a new random variable $X+2$ . Its mean would be

$E[X+2]=E[X]+E[2]=E[X]+2$

since expectation is linear, and the expected value of a constant is that constant. $E[X]$ is the mean of $X$ , so the new mean would be

$E[X+2]=10+2=12$

The variance of a random variable $X$ is

$V[X]=E[X^2]-E[X]^2$

so the variance of $X+2$ would be

$V[X+2]=E[(X+2)^2]-E[X+2]^2$

We already know $E[X+2]=12$ , so simplifying above, we get

$V[X+2]=E[X^2+4X+4]-12^2$

$V[X+2]=E[X^2]+4E[X]+4-12^2$

$V[X+2]=(V[X]+E[X]^2)+4E[X]-140$

Standard deviation is the square root of variance, so $V[X]=3^2=9$ .

$\implies V[X+2]=(9+10^2)+4(10)-140=9$

so the standard deviation remains unchanged at 3.

NB: More generally, the variance of $aX+b$ for $a,b\in\mathbb R$ is

$V[aX+b]=a^2V[X]+b^2V[1]$

but the variance of a constant is 0. In this case, $a=1$ , so we're left with $V[X+2]=V[X]$ , as expected.

5 0

3 years ago