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3 years ago

Each day that a library book is kept past its due date, a $0.30 fee is charged at midnight. Which ordered pair is

1 answer:

mihalych1998 [28]3 years ago

6 0

Let's say the number of days library book is late is X, and the total fee is Y.

Liability charges $0.30 dollars as a fee for being 1 day late,

For being 1 day late, fee charge is: 1* $0.30
So, for X days the charge would be: X*$0.30.

Total charge for being X days late is Y, Which means: Y= 0.30 * X.

Now We would have to check all the viable solutions in the answer to see if they satisfy the equation Y= 0.30 * X

Option one(-3, -0.9) and two (-2.5, -0.75) Would not be a viable solution because the value of number of days can not be negative and in option one and two, value of days -3 and -2.5 is negative.

Option three(4.5, 1.35) can not be correct because library charges fee for a full day so the number for days would be a whole number. Library would not charge for 4.5 days, they would either charge of 4 days or 5 days because 4.5 is not an whole number.

Option four(8, 2.40) is the correct answer because it satisfies our equation;

Y= 0.30 * X

2.40= 0.30 * 8

2.40 = 2.40.

Fourth option (8, 2.40) is the only viable solution to this question.

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3 years ago

An aircraft on a reconnaissance mission takes off from its home base and flies 550 miles at a bearing of S 46° E to a location i

Alexandra [31]

<h3>Answer: D) 1692 miles</h3>

============================================

Explanation:

A diagram is very handy for this type of problem. Refer to the diagram below (attached image). Here are the steps used to form the diagram

Plot point A and make that the home base. Plot point D directly south of point A. I'm skipping B and C because those will be used to form the triangle later.
Imagine you are standing at point A and looking south at point D. Turn 46 degrees toward the east direction (this is what the notation "bearing of S 46° E" means). Then move 550 miles from A to land on point B. This is the location where the plane makes its first turn.
Plot the points E and F, which are north and south of point B.
Imagine we are at point B and looking directly south at point F. Turn 55 degrees toward the west, which is due to the "S 55° W" and then move 483 miles to arrive at point C. This is the plane's location where it turns around to go back to home base.

The diagram below shows all of this visually summarized. We have the segments

AB = 550 miles
BC = 483 miles
AC = x miles

And we have these angles

Angle DAB = 46 degrees and Angle ABE = 46 degrees (blue)
Angle ABC = 79 degrees (red)
Angle ACB = 55 degrees and Angle FBC = 55 degrees (green)

Angle ABC comes from the fact that the blue and green angles add with the red angle to get 180. So we basically need to solve 46+y+55 = 180 which leads to y = 79.

------------------------------

Once the diagram is set up, we'll use the law of cosines to find x.

Focus on triangle ABC. Ignore points D,E,F. Ignore any extra lines as well.

The interior angles of triangle ABC are

A = 46 degrees
B = 79 degrees
C = 55 degrees

The sides opposite the angles are

a = 483 (opposite angle A)
b = x (opposite angle B)
c = 550 (opposite angle C)

Now we can apply the law of cosines

b^2 = a^2 + c^2 - 2*a*c*cos(B)

x^2 = 483^2 + 550^2 - 2*483*550*cos(79)

x^2 = 434412.180756441

x = sqrt(434412.180756441)

x = 659.099522649229

x = 659

Side AC, or CA, is roughly 659 miles long.

------------

To find the total distance the plane travels, we add up the three sides of the triangle (ie we find the perimeter)

Distance Traveled = AB + BC + CA

Distance Traveled = 550 + 483 + 659

Distance Traveled = 1692 miles

3 0

3 years ago

Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

So the two square roots of z are

$\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

and

$\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

3 0

3 years ago

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