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3 years ago

13

Which is the domain of the function f(x) = -5/6(3/5)^x?

1 answer:

lutik1710 [3]3 years ago

7 0

the domain of f is ${\bf R} = (-\infty, \infty)$

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Find all the critical values

Natalija [7]

y - 3
g(y) = ------------------
y^2 - 3y + 9

To find the c. v., we must differentiate this function g(y) and set the derivative equal to zero:

(y^2 - 3y + 9)(1) - (y - 3)(2y - 3)
g '(y) = --------------------------------------------
(y^2 - 3y + 9)^2

Note carefully: The denom. has no real roots, so division by zero is not going to be an issue here.

Simplifying the denominator of the derivative,

(y^2 - 3y + 9)(1) - (y - 3)(2y - 3) => y^2 - 3y + 9 - [2y^2 - 3y - 6y + 9], or
-y^2 + 6y

Setting this result = to 0 produces the equation y(-y + 6) = 0, so
y = 0 and y = 6. These are your critical values. You may or may not have max or min at one or the other.

6 0

2 years ago

Bookstore sells book in excellent condition for $2.50 and in fair condition for $0.50 . Write an expression for cost of buying x

alexdok [17]

X = excellent condition books
y = fair condition books

2.50x + 0.50y <== ur expression

8 0

2 years ago

Hya think of a number "m" . She square it, add "y" to the answer and then subtract 7 from it. The final answer is "A". 1) Derive

Bond [772]

<h2><u>Solution (1)</u> :</h2>

Given, to find A we have to :

square m
Add y to m²
Subtract 7 from m² + y

From the question, the following equation can be formed :

$=\tt A = {m}^{2} + y - 7$

Therefore, the formula for finding A = m² + y - 7

<h2><u>Solution (2)</u> :</h2>

The value of A we can derive from the formula is :

$=\tt A = {m}^{2} + y - 7$

Value of m = 3 (given)

Which means :

$=\tt A = {3}^{2} + y - 7$

$=\tt \: A = 9 + y - 7$

$=\tt A = 9 - 7 + y$

$\color{plum} =\tt A = 2 + y$

Thus, the value of A = 2+y

Therefore, the value of A = <u>2+y</u>

7 0

2 years ago

Which of the following points is a solution of the inequality y < -|x|?

Rzqust [24]

y < -|x|

replace the letters with the given numbers:

(1,-2) -2<-|1| this is true

(1,-1) -1 <-|1| this is false

(1,0) 0 < -|1| this I false

The answer is (1,-2)

5 0

2 years ago

Read 2 more answers

Find the smallest positive $n$ such that \begin{align*} n &\equiv 3 \pmod{4}, \\ n &\equiv 2 \pmod{5}, \\ n &\equiv

Alex777 [14]

4, 5, and 7 are mutually coprime, so you can use the Chinese remainder theorem right away.

We construct a number $x$ such that taking it mod 4, 5, and 7 leaves the desired remainders:

$x=3\cdot5\cdot7+4\cdot2\cdot7+4\cdot5\cdot6$

Taken mod 4, the last two terms vanish and we have

$x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod4$

so we multiply the first term by 3.

Taken mod 5, the first and last terms vanish and we have

$x\equiv4\cdot2\cdot7\equiv51\equiv1\pmod5$

so we multiply the second term by 2.

Taken mod 7, the first two terms vanish and we have

$x\equiv4\cdot5\cdot6\equiv120\equiv1\pmod7$

so we multiply the last term by 7.

Now,

$x=3^2\cdot5\cdot7+4\cdot2^2\cdot7+4\cdot5\cdot6^2=1147$

By the CRT, the system of congruences has a general solution

$n\equiv1147\pmod{4\cdot5\cdot7}\implies\boxed{n\equiv27\pmod{140}}$

or all integers $27+140k$ , $k\in\mathbb Z$ , the least (and positive) of which is 27.

3 0

3 years ago