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3 years ago

Can someone tell me which one it is

Mathematics

1 answer:

IrinaK [193]3 years ago

3 0

Answer:

Step-by-step explanation:

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A hockey player strikes a hockey puck. The height of the puck increases until it reaches a maximum height of 3 feet, 55 feet awa

tatuchka [14]

Answer:

Second puck travels farther

Step-by-step explanation:

Maximum height of first puck = 3 feet

The height of a second hockey puck is modeled by:

$y=x\left(0.15-0.001x\right)$

$y=0.15x-0.001x^2$

To find maximum height of second puck

$y'=0.15-0.002x$

Equate the derivative equals to 0

0.15-0.002x=0

$\frac{0.15}{0.002}=x$

75 = x

At x = 75

$y=0.15(75)-0.001(75)^2=5.625$

So, The maximum height of second puck is greater than first puck

So, Second puck travels farther

5 0

3 years ago

Select the correct answer.

dimulka [17.4K]

Answer:

A, y=-6

Step-by-step explanation

Because it isssssssssssssssssssssssssss a watermelon

3 0

3 years ago

Select all the correct answers.<br> Which statements are true about function g?

erica [24]

Answer:

i need the question

Step-by-step explanation:

8 0

2 years ago

How many 0.25 litre cups can be filled from a 4.51 jug of lemonade

siniylev [52]

To find out the cups, total jug volume should be divided by 0.25 litre.
4.5/0.25
18 cups

8 0

3 years ago

Sec squared 55 - tan squared 55

sergejj [24]

<u>Answer: </u>

sec squared 55 – tan squared 55 = 1

<u>Explanation:</u>

Given, sec square 55 – tan squared 55

We know that,

$\sec \Theta=\frac{\text {hypotenuse}}{\text {base}}$

And,

$\tan \theta=\frac{\text { perpendicular }}{\text { base }}$

where Ө is the angle

Substituting the values

$\left(\frac{\text {hypotenuse}}{\text {base}}\right)^{2}-\left(\frac{\text { perpendicular }}{\text {base}}\right)^{2}$

Solving,

$\frac{(\text {hypotenuse})^{2}-(\text {perpendicular})^{2}}{(\text {base}) *(\text {base})}$

According to Pythagoras theorem,

$\text { (hypotenuse) }^{2}-\text { (perpendicular) }^{2}=(\text { base })^{2}$

Putting this in the equation;

squared 55 - tan squared 55 =

$\frac{(\text {hypotenuse})^{2}-(\text {perpendicular})^{2}}{(\text {base}) *(\text {base})}=\frac{(\text {base})^{2}}{(\text {base}) *(\text {base})}=1$

Therefore, sec squared 55 – tan squared 55 = 1

6 0

3 years ago