All categories

2 years ago

What are true of the function f(x)=49(1/7)x? Check all that apply

Mathematics

2 answers:

ahrayia [7]2 years ago

7 0

The given function is : y= f(x) = $49 [\frac{1}{7}]^x$

The statements which are true regarding this function is

1. As you can observe that y is defined for all values of x , so Domain is set of all values that x can take which is set of all real numbers.

2. Also, $x =\frac {\log[\frac{y}{49}]}{-\log7}$ , so x is defined for all values of y. Range is all values that y can take which is also set of all real numbers greater than zero, i.e y>0.

As ,x =0 then y= 49 and $x_{1}$ =1 then $y_{1}$ = 7,

so , $y_{1} = \frac{y}{7}$ .

So, Option (1),(4) and (5) are correct.

VLD [36.1K]2 years ago

5 0

The domain is all real numbers
The Range is y > 0
AS x increases by 1 the value of y is 1/7 of previous value

You might be interested in

A cube has a side length of 8 inches.

Kazeer [188]

The answers are $8^{3}$ and 8 · 8 · 8

Explanation:

The volume of a cube or space a cube occupies is calculated by multiplying the length, height, and width. Additionally, because in a cube all sides are equal in length, the formula for volume is V= $s^{3}$ . According to this, if the side length is 8 inches the volume can be represented as $V = 8^{3}$ or V= 8 · 8 · 8 because the symbol · represents multiplication. Also, this can be solved as V = 512 cubic inches. This makes the following options correct: $8^{3}$ and 8 · 8 · 8.

4 0

3 years ago

Can someone provide a step by step explanation on how to solve? Cheers.

blsea [12.9K]

Answer:

a) 0.0137060668

please mark me brainliest and follow me my friend.

I not know b part.

6 0

2 years ago

BRAINLIEST AWARD NO.2

zzz [600]

Answer:

A whole number first term to render as fifth term a value larger than 10000, should be at least 121

Step-by-step explanation:

The formula is given as recursive since it involves the previous number of the sequence, and defined as:

$a_n=a_{n-1}*3+6$

we also know that the first term is 4

Then in this case, the first five terms are:

$a_1=4\\a_2=4*3+6=18\\a_3=18*3+6=60\\a_4=60*3+6=186\\a_5=186*3+6=564\\$

So if we want to find the first term in the case that the fifth one is greater than 10,000 using this recursive formula, now we have to start backwards, and say that the fifth term is "> 10000" and what the fourth one is.

Notice that if you have this definition for the nth term, we can obtain from it, what the previous term is to find the general rule:

$a_n=a_{n-1}*3+6\\a_n-6=a_{n-1}*3\\\frac{a_n-6}{3} = a_{n-1}\\a_{n-1}=\frac{a_n}{3} -2$

So the rule is to subtract 6 from he term, and divide the subtraction by 3. Then working backwards:

$a_5>10000\\\frac{a_5}{3} -2>\frac{10000}{3} -2\\a_4>=\frac{10000}{3} -2\\\frac{a_4}{3} -2>\frac{\frac{10000}{3}-2}{3}-2 =\frac{10000}{9}-\frac{8}{3} \\a_3>\frac{10000}{9}-\frac{8}{3} \\\frac{a_3}{3} -2>\frac{\frac{10000}{9}-\frac{8}{3} }{3} -2=\frac{10000}{27} -\frac{8}{9} -2=\frac{10000}{27} -\frac{26}{9}\\a_2=\frac{10000}{27} -\frac{26}{9}\\\frac{a_2}{3} -2>\frac{\frac{10000}{27} -\frac{26}{9}}{3} -2=\frac{10000}{81} -\frac{80}{27} \\a_1>\frac{10000}{81} -\frac{80}{27}\approx 120.49$