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Lorico [155]
4 years ago
11

A woman bought a horse for $800 and then sold it for $900. She bought it back for $400 and then sold it again for $600. How much

did she gain or lose on these transactions?
Mathematics
1 answer:
spayn [35]4 years ago
8 0
1st- gained lost 800, gained 900
2- lost 400- gained 600

so in all she lost 1200
but then gained 1500
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13. f(x) = x + 5<br> a. f(4)<br> b. f(7)<br> c. f(-3)<br> d. f(0)<br> e. f(2.4)<br> f. f(2/3)
SashulF [63]
A.9
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C.2
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The length of a rectangle is twice its width.
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8 0
3 years ago
Which function has the greatest rate of change on the interval from x = 3 pi over 2 to x = 2π?
vodka [1.7K]
The average rate of change for the function f(x) can be calculated from the following equation
\frac{f( x_{2})-f( x_{1} )}{ x_{2} - x_{1} }

By applying the last formula on the given equations 
(1) the first function f
from the table f(3π/2) = -2   and   f(2π) = 0
∴ The average rate of f = \frac{f(2 \pi)-f( \frac{3 \pi}{2} )}{2 \pi -  \frac{3 \pi}{2} } =  \frac{0-(-2)}{ \frac{\pi}{2} }=  \frac{2}{ \frac{\pi}{2} }  =  \frac{4}{\pi}

(2) the second function g(x)
from the graph g(3π/2) = -2   and   g(2π) = 0
∴ The average rate of g = \frac{g(2 \pi)-g( \frac{3 \pi}{2} )}{2 &#10;\pi -  \frac{3 \pi}{2} } =  \frac{0-(-2)}{ \frac{\pi}{2} }=  \frac{2}{ &#10;\frac{\pi}{2} }  =  \frac{4}{\pi}

(3) the third function h(x) = 6 sin x +1
∴ h(3π/2) = 6 sin (3π/2) + 1 = 6 *(-1) + 1 = -5
   h(2π) = 6 sin (2π) + 1 = 6 * 0 + 1 = 1
∴ The average rate of h = \frac{f(2 \pi)-f( \frac{3 \pi}{2} )}{2 &#10;\pi -  \frac{3 \pi}{2} } =  \frac{1-(-5)}{ \frac{\pi}{2} }=  \frac{6}{ &#10;\frac{\pi}{2} }  =  \frac{12}{\pi}

By comparing the results, The <span>function which has the greatest rate of change is h(x)
</span>

So, the correct answer is option <span>C) h(x)</span>
4 0
3 years ago
All three please 10 points
GaryK [48]

Answer:

A) x = 3 or -1

B) x = -7

C)x = -7

Step-by-step explanation:

A) x² + 2x + 1 = 2x² - 2

Rearranging, we have;

2x² - x² - 2x - 2 - 1 = 0

x² - 2x - 3 = 0

Using quadratic formula, we have;

x = [-(-2) ± √((-2)² - 4(1 × -3))]/(2 × 1)

x = (2 ± √16)/2

x = (2 + 4)/2 or (2 - 4)/2

x = 6/2 or -2/2

x = 3 or -1

B) ((x + 2)/3) - 2/15 = (x - 2)/5

Multiply through by 15 to get;

5(x + 2) - 2 = 3(x - 2)

5x + 10 - 2 = 3x - 6

5x - 3x = -6 - 10 + 2

2x = -14

x = -14/2

x = -7

C) log(2x + 3) = 2log x

From log derivations, 2 log x is same as log x²

Thus;

log(2x + 3) = logx²

Log will cancel out to give;

2x + 3 = x²

x² - 2x - 3 = 0

Using quadratic formula, we have;

x = [-(-2) ± √((-2)² - 4(1 × -3))]/(2 × 1)

x = (2 ± √16)/2

x = (2 + 4)/2 or (2 - 4)/2

x = 6/2 or -2/2

x = 3 or -1

5 0
3 years ago
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