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Angelina_Jolie [31]
3 years ago
9

A fish jumps out of the water at a speed of 12 feet per second. The height y (in feet) of the fish above the surface of the wate

r is represented by the equation y=-16x^2+12x, where x is the time (in seconds) since the jump began. The fish reaches its highest point above the surface of the water after 0.375 seconds. How far above the surface is the fish at this time?
Mathematics
1 answer:
sasho [114]3 years ago
8 0

Answer:

The fish is 2.25 ft above the surface at 0.375 seconds

Step-by-step explanation:

Given:

y=-16x^2+12x

x=0.375 seconds

Substitute x=0.375 into the equation

y=-16x^2+12x

= -16(0.375)^2 +12(0.375

= -16(0.140625) + 4.5

= -2.25 + 4.5

= 2.25

y=2.25 ft

The fish is 2.25 ft above the surface at 0.375 seconds

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If y varies inversely as x, find the constant of variation if y=36 when x=9
Jobisdone [24]

Answer:

k = 324

Step-by-step explanation:

Y varies inversely as x

Let k be the constant of variation

Y = k/ x

Y= 36 and x= 9

Substitute the value of Y and x

36 = k/9

Make k the subject of the formula

k= 36*9

k =324

5 0
3 years ago
Find 2x2 − 2z4 + y2 − x2 + z4 if x = −4, y = 3, and z = 2.
maw [93]
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After that it's simple addition giving you an answer of -26.
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3 0
3 years ago
Find the 2nd term of (2x+3y)^3
puteri [66]

Answer:

The 2nd term is 3y

Step-by-step explanation:

2x +       3y

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2 years ago
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Working at home: According to the U.S Census Bureau, 41% of men who worked at home were college graduates. In a sample of 506 wo
sergejj [24]

Solution :

a). The point estimate of proportion of college graduates among women who work at home,

$\hat p =\frac{166}{506}$

  = 0.3281

99.5% confidence interval

$=\left( \hat p \pm Z_{0.005/2} \sqrt{\frac{\hat p (1- \hat p)}{n}} \right)$

$=\left( 0.3281 \pm 2.81 \sqrt{\frac{0.3281 \times  (1- 0.3281)}{506}} \right)$

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3 years ago
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Nesterboy [21]
The answer is D!!!!!
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3 years ago
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