All categories

2 years ago

Which of the functions below could have created this graph?

Mathematics

2 answers:

monitta2 years ago

6 0

Answer:

I think A

Step-by-step explanation:

iren [92.7K]2 years ago

5 0

Answer:

D. -1/2x^4-x^3+x+2

Step-by-step explanation:

plug in the graphing calculator.

You might be interested in

(1 point) There were 23.7 million licensed drivers in California in 2009 and 22.76 million in 2004. Find a formula for the numbe

Salsk061 [2.6K]

Answer:

N(t) = 0.188t + 22.76

Step-by-step explanation:

Number of licensed drivers in 2004 = 22.76 million

Number of licensed drivers in 2009 = 23.7 million

Number of licensed drivers, N as a function of t since year 2004 ;

General form of a linear function :

y = mx + c

c = intercept ; m = slope

Intercept c = value of y ; when x = 0

Here, population after uerssmmx,

Hence,

In 2004 ;

22.76 = mx + c

x = 0

22.76 = c

Number in 2009

x = number of yesrs after 2004 ; x = 2000 - 2004 = 5years

We can find the slope :

y = m*5 + 22.76

y = 23.7 in 2009

23.7 = 5m + 22.76

23.7 - 22.76 = 5m

m = 0.94 / 5

m = 0.188

Hence, the linear function can be written as :

N(t) = 0.188t + 22.76

8 0

3 years ago

WHAT IS THE REMAINDER WHEN <img src="https://tex.z-dn.net/?f=32%5E%7B37%5E%7B32%7D%20%7D" id="TexFormula1" title="32^{37^{32} }"

Feliz [49]

Recall Euler's theorem: if $\gcd(a,n) = 1$ , then

$a^{\phi(n)} \equiv 1 \pmod n$

where $\phi$ is Euler's totient function.

We have $\gcd(9,32) = 1$ - in fact, $\gcd(9,32^k)=1$ for any $k\in\Bbb N$ since $9=3^2$ and $32=2^5$ share no common divisors - as well as $\phi(9) = 6$ .

Now,

$37^{32} = (1 + 36)^{32} \\\\ ~~~~~~~~ = 1 + 36c_1 + 36^2c_2 + 36^3c_3+\cdots+36^{32}c_{32} \\\\ ~~~~~~~~ = 1 + 6 \left(6c_1 + 6^3c_2 + 6^5c_3 + \cdots + 6^{63}c_{32}\right) \\\\ \implies 32^{37^{32}} = 32^{1 + 6(\cdots)} = 32\cdot\left(32^{(\cdots)}\right)^6$

where the $c_i$ are positive integer coefficients from the binomial expansion. By Euler's theorem,

$\left(32^{(\cdots)\right)^6 \equiv 1 \pmod9$

so that

$32^{37^{32}} \equiv 32\cdot1 \equiv \boxed{5} \pmod9$

7 0

1 year ago

Please hellllllllppppppppppppppppppppppppppp

vitfil [10]

Answer:

$21^{10}$

Step-by-step explanation:

First, multiply the 3 and the 7 together, and then "multiply" the exponents -8 and 3. (However, when "multiplying exponents, you're really just adding them together: -8 + 5)

( $21^{-5}$ )^2