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bearhunter [10]
3 years ago
7

A deep sea diver is at 45 feet below sea level. A helicopter is at 100 feet above sea level. What is the distance between the he

licopter and deep sea diver?
Mathematics
1 answer:
Luden [163]3 years ago
5 0

Answer:

D = 55 feet

Step-by-step explanation:

Let below sea level is negative direction and above sea level is positive direction.

A deep sea diver is at 45 feet below sea level = -45 feet

A helicopter is at 100 feet above sea level = +100 feet

We need to find the distance between the helicopter and deep sea diver. It can be calculated by simply adding the above data as follows:

D = -45 +100

= 55 feet

Hence, the distance between the helicopter and deep sea diver is 55 feet.

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Consider the Polynomial P(x)=x^3-5x^2-x+5. is (x-5) a factor?
Alborosie

Answer:

Option B, Yes, the remainder is 0, so x-5 is a factor of P(x)

Step-by-step explanation:

<u>Step 1:  Factor</u>

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

<em>p(x) = (x - 5)(x + 1)(x - 1)</em>

<em />

<em>Yes, the remainder is 0 so, x-5 is a factor of p(x)</em>

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Answer:  Option B, Yes, the remainder is 0, so x-5 is a factor of P(x)

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3 years ago
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The height H of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 9
mariarad [96]

Answer:

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

Step-by-step explanation:

Statement is incorrect. Correct form is presented below:

<em>The height </em>h(t)<em> of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 90 feet per second is given by equation </em>h(t) = 3 +90\cdot t -16\cdot t^{2}<em>, where </em>t<em> is time in seconds. After how many seconds will the ball be 84 feet above the ground. </em>

We equalize the kinematic formula to 84 feet and solve the resulting second-order polynomial by Quadratic Formula to determine the instants associated with such height:

3+90\cdot t -16\cdot t^{2} = 84

16\cdot t^{2}-90\cdot t +81 = 0 (1)

By Quadratic Formula:

t_{1,2} = \frac{90\pm \sqrt{(-90)^{2}-4\cdot (16)\cdot (81)}}{2\cdot (16)}

t_{1} = 4.5\,s, t_{2} = 1.125\,s

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

3 0
3 years ago
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Answer:

18

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12 and 15 are both dividable by 3 and 3 apart so if you subtract 21 by 3 than you get 18

7 0
2 years ago
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goblinko [34]

Answer:

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