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kipiarov [429]
3 years ago
10

What is y+2=⅞(x-3) in standard form?

Mathematics
1 answer:
Alla [95]3 years ago
8 0
The answer is 7x - 8y = 37
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Hi, am I supposed to find the stationary points of the f(x) and g(x) then use the distance formula to solve for the final answer
nikdorinn [45]

9514 1404 393

Answer:

  maximum difference is 38 at x = -3

Step-by-step explanation:

This is nicely solved by a graphing calculator, which can plot the difference between the functions. The attached shows the maximum difference on the given interval is 38 at x = -3.

__

Ordinarily, the distance between curves is measured vertically. Here that means you're interested in finding the stationary points of the difference between the functions, along with that difference at the ends of the interval. The maximum difference magnitude is what you're interested in.

  h(x) = g(x) -f(x) = (2x³ +5x² -15x) -(x³ +3x² -2) = x³ +2x² -15x +2

Then the derivative is ...

  h'(x) = 3x² +4x -15 = (x +3)(3x -5)

This has zeros (stationary points) at x = -3 and x = 5/3. The values of h(x) of concern are those at x=-5, -3, 5/3, 3. These are shown in the attached table.

The maximum difference between f(x) and g(x) is 38 at x = -3.

4 0
3 years ago
A page was missing from a 72 - page book. When Josh added up the remaining page numbers, the sum was 2521. Which page was missin
morpeh [17]
The numbers 1 to 72 add up to (1+72) x 72/2 = 73 x 36 = 2628
If we take away what Josh got we have 2628 - 2521 = 107
Pages in a book are odd on the first side and even on the second, so pages 11 and 12 are on the same sheet.
107 / 2 = 53.5 so the missing page had 53 on one side and 54 on the other side.
The remaining pages would have been numbered 1, 2, 3, ..., 51, 52, 55, 56, ... 71, 72
8 0
3 years ago
Find the sum of the infinite series 1/3+4/9+16/27+64/81+.... if it exists.
pshichka [43]
9/27, 12/27, 16/27

So this is a geometric sequence as each term is 4/3 the previous term.

Since the common ration is greater than one the sum of the series diverges, it does not exist.  (The sum just keeps getting larger and larger)

For a geometric series to have a sum r^2<1

So that the normal sum....

s(n)=a(1-r^n)/(1-r)   becomes if r^2<1

s=a/(1-r)
6 0
3 years ago
What is the zero of function f?f(x)=3 square root of x+3 -6
grin007 [14]

Solution:

Given:

f(x)=3\sqrt{x+3}-6

The zeros of a function are the values of x when f(x) is equal to 0.

Hence,

\begin{gathered} 0=3\sqrt{x+3}-6 \\  \\ Collecting\text{ the like terms,} \\ 0+6=3\sqrt{x+3} \\ 6=3\sqrt{x+3} \\  \\ Divide\text{ both sides by 3;} \\ \frac{6}{3}=\sqrt{x+3} \\ 2=\sqrt{x+3} \\  \\ Taking\text{ the square of both sides;} \\ 2^2=x+3 \\ 4=x+3 \\  \\ Collecting\text{ the like terms;} \\ 4-3=x \\ 1=x \\ x=1 \end{gathered}

Therefore, x = 1

The correct answer is OPTION A.

3 0
1 year ago
F(x) = 3x - 2x - 5<br><br> Factoring quadratics
bagirrra123 [75]

Answer:

(3x -5)(x+1)

Step-by-step explanation:

Factoring 3x^{2} -2x-5=0 results in (3x -5)(x+1)

4 0
2 years ago
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