Can I get some help on this, please??

Mathematics

1 answer:

sdas [7]2 years ago

6 0

Answer:

0.00501, the -3 means to move the decimal point to the left three

Hope this helps ;)

Step-by-step explanation:

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Imagine that you are standing at the top of a cliff that is 1,500 feet above sea level. You see ship A, and your eyes make an an

kati45 [8]

Answer: option A. 30 °.

Explanation:

1) Draw this triangle: Eyes - Ship A - height of the cliff:

The angle of the vertix where the eyes are is: 90° - depreesion angle = 90° - 60° = 30°.

2) opossite leg = horizontal distance to the ship A = x

3) adjacent leg = height of the cliff = 1500 feet

4) tan(30°) = x / 1500 feet => x = 1500 tan(30°) = 1500 √3 / 3 = 500√3

5) Draw the triangle of the eyes - ship B - hieght of the cliff

6) The unknown is the depression angle of the eyes with ship B, α

7) opposite leg = 500√3 feet + 1000√3 feet = 1500√3 feet

8) adjacent leg = height of the cliff = 1500 feet

9) tan(α) = opposite leg / adjacent leg = 1500√3 / 1500 = √3

10) α = arctan(√3) = 60°

11) Depression angle = 90° - 60° = 30° Which is the answer

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3 years ago

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What is the completely factored form of d4 -8d2 + 16

irinina [24]

According to Vieta's Formulas, if $x_1,x_2$ are solutions of a given quadratic equation:

$ax^2+bx+c=0$

Then:

$a(x-x_1)(x-x_2)$ is the completely factored form of $ax^2+bx+c$ .

If choose $x=d^2$ , then:

$\displaystyle x^2-8x+16=0\\\\x_{1,2}= \frac{8\pm \sqrt{64-64} }{2}=4$

So, according to Vieta's formula, we can get:

$x^2-8x+16=(x-4)(x-4)= (x-4)^2$

But $x=d^2$ :

$d^4-8d^2+16=(d^2-4)^2=[(d+2)(d-2)]^2=(d+2)^2(d-2)^2$

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2 years ago

The population of a town increased by 10% per year in 2009, 2010, and 2011. If the population of the town at the beginning of 20

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If the population was 40,000 at the beginning of 2009, at the end of 2009 it has increased by 10% and became 44,000.

By the same way, at the end of 2010, it became $44000 + \frac{44000*10}{100} = 48400$

At the end of 2011, the population was $48400 + \frac{48400 * 10}{100} = 53240$