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3 years ago

8

How do you simplify this?

3i \sqrt{ - 12} " alt=" - 3i \sqrt{ - 12} " align="absmiddle" class="latex-formula">

2 answers:

ohaa [14]3 years ago

7 0

You simplify the square root -12 to 2i square root 3
You get -6i^2 square root 3
And if you simplify it you get 6 square root 3

Mariana [72]3 years ago

5 0

Since you can think of i as the square root of -1, we have

$\sqrt{-12} = \sqrt{12\cdot(-1)} = \sqrt{12}\sqrt{-1} = 12i$

So, you have

$-3i\sqrt{-12}=-3i(12i)=-36i^2=36$

You might be interested in

Ally's tricycle has two different wheel sizes: one larger wheel and two smaller wheels. As Ally travels the distance between her

vazorg [7]

Answer:

500 inches

Step-by-step explanation:

Let C₁ and n₁ be the circumference and the number of rotations of the larger wheel. Also, let C₂ and n₂ be the circumference and the number of rotations of the smaller wheel. Since the distance moved by both wheels from Ally's house to her neighbor's house is the same, n₁C₁ = n₂C₂. (1)

Also it is given that half the circumference of the larger wheel is 5 inches more than the circumference of one of the smaller wheels.

So, C₁/2 = C₂ + 5

C₁ = 2(C₂ + 5) (2)

Substituting C₁ into (1), we have

n₁C₁ = n₂C₂.

n₁[2(C₂ + 5)] = n₂C₂.

Since n₁ = 10 and n₂ = 25, we have

10[2(C₂ + 5)] = 25C₂

20(C₂ + 5) = 25C₂

expanding the bracket,

20C₂ + 100 = 25C₂

collecting like terms, we have

25C₂ - 20C₂ = 100

5C₂ = 100

dividing both sides by 5, we have

C₂ = 100/5

= 20 inches

So, the distance between Ally's house and her neighbor's house is d = n₂C₂ = 25(20)

= 500 inches

8 0

3 years ago

6 math questions, answer all please for all points

Sergeeva-Olga [200]

Answer:

See below for answers and explanations

Step-by-step explanation:

<u>Problem 1</u>

Recall that the projection of a vector $u$ onto $v$ is $\displaystyle proj_vu=\biggr(\frac{u\cdot v}{||v||^2}\biggr)v$ .

Identify the vectors:

$u=\langle-10,-7\rangle$

$v=\langle-8,4\rangle$

Compute the dot product:

$u\cdot v=(-10*-8)+(-7*4)=80+(-28)=52$

Find the square of the magnitude of vector v:

$||v||^2=\sqrt{(-8)^2+(4)^2}^2=64+16=80$

Find the projection of vector u onto v:

$\displaystyle proj_vu=\biggr(\frac{u\cdot v}{||v||^2}\biggr)v\\\\proj_vu=\biggr(\frac{52}{80}\biggr)\langle-8,4\rangle\\\\proj_vu=\biggr\langle\frac{-416}{80} ,\frac{208}{80}\biggr\rangle\\\\proj_vu=\biggr\langle\frac{-26}{5} ,\frac{13}{5}\biggr\rangle\\\\proj_vu=\langle-5.2,2.6\rangle$

Thus, B is the correct answer

<u>Problem 2</u>

Treat the football and wind as vectors:

Football: $u=\langle42\cos172^\circ,42\sin172^\circ\rangle$

Wind: $v=\langle13\cos345^\circ,13\sin345^\circ\rangle$

Add the vectors: $u+v=\langle42\cos172^\circ+13\cos345^\circ,42\sin172^\circ+13\sin345^\circ\rangle\approx\langle-29.034,2.481\rangle$

Find the magnitude of the resultant vector:

$||u+v||=\sqrt{(-29.034)^2+(2.481)^2}\approx29.14$

Find the direction of the resultant vector:

$\displaystyle \theta=tan^{-1}\biggr(\frac{2.841}{-29.034}\biggr)\approx -5^\circ$

Because our resultant vector is in Quadrant II, the true direction angle is 6° clockwise from the negative axis. This means that our true direction angle is $180^\circ-5^\circ=175^\circ$

Thus, C is the correct answer

<u>Problem 3</u>

We identify the initial point to be $R(-2,12)$ and the terminal point to be $S(-7,6)$ . The vector in component form can be found by subtracting the initial point from the terminal point:

$v=\langle-7-(-2),6-12\rangle=\langle-7+2,-6\rangle=\langle-5,-6\rangle$

Next, we find the magnitude of the vector:

$||v||=\sqrt{(-5)^2+(-6)^2}=\sqrt{25+36}=\sqrt{61}\approx7.81$

And finally, we find the direction of the vector:

$\displaystyle \theta=tan^{-1}\biggr(\frac{6}{5}\biggr)\approx50.194^\circ$

Keep in mind that since our vector is in Quadrant III, our direction angle also needs to be in Quadrant III, so the true direction angle is $180^\circ+50.194^\circ=230.194^\circ$ .

Thus, A is the correct answer

<u>Problem 4</u>

Add the vectors:

$v_1+v_2=\langle-60,3\rangle+\langle4,14\rangle=\langle-60+4,3+14\rangle=\langle-56,17\rangle$

Determine the magnitude of the vector:

$||v_1+v_2||=\sqrt{(-56)^2+(17)^2}=\sqrt{3136+289}=\sqrt{3425}\approx58.524$

Find the direction of the vector:

$\displaystyle\theta=tan^{-1}\biggr(\frac{17}{-56} \biggr)\approx-17^\circ$

Because our vector is in Quadrant II, then the direction angle we found is a reference angle, telling us the true direction angle is 17° clockwise from the negative x-axis, so the true direction angle is $180^\circ-17^\circ=163^\circ$

Thus, A is the correct answer

<u>Problem 5</u>

A vector in trigonometric form is represented as $w=||w||(\cos\theta i+\sin\theta i)$ where $||w||$ is the magnitude of vector $w$ and $\theta$ is the direction of vector $w$ .

Magnitude: $||w||=\sqrt{(-16)^2+(-63)^2}=\sqrt{256+3969}=\sqrt{4225}=65$

Direction: $\displaystyle \theta=tan^{-1}\biggr(\frac{-63}{-16}\biggr)\approx75.75^\circ$

As our vector is in Quadrant III, our true direction angle will be 75.75° counterclockwise from the negative x-axis, so our true direction angle will be $180^\circ+75.75^\circ=255.75^\circ$ .

This means that our vector in trigonometric form is $w=65(\cos255.75^\circ i+\sin255.75^\circ j)$

Thus, C is the correct answer

<u>Problem 6</u>

Write the vectors in trigonometric form:

$u=\langle40\cos30^\circ,40\sin30^\circ\rangle\\v=\langle50\cos140^\circ,50\sin140^\circ\rangle$

Add the vectors:

$u+v=\langle40\cos30^\circ+50\cos140^\circ,40\sin30^\circ+50\sin140^\circ\rangle\approx\langle-3.661,52.139\rangle$

Find the magnitude of the resultant vector:

$||u+v||=\sqrt{3.661^2+52.139^2}\approx52.268$

Find the direction of the resultant vector:

$\displaystyle\theta=tan^{-1}\biggr(\frac{52.139}{-3.661} \biggr)\approx-86^\circ$

Because our resultant vector is in Quadrant II, then our true direction angle will be 86° clockwise from the negative x-axis. So, our true direction angle is $180^\circ-86^\circ=94^\circ$ .

Thus, B is the correct answer

5 0

2 years ago

In how many ways can we seat 3 pairs of siblings in a row of 7 chairs, so that nobody sits next to their sibling

monitta

Answer:

1,968

Step-by-step explanation:

Let x₁ and x₂, y₁ and y₂, and z₁ and z₂ represent the 3 pairs of siblings, and let;

Set X represent the set where the siblings x₁ and x₂ sit together

Set Y represent the set where the siblings y₁ and y₂ sit together

Set Z represent the set where the siblings z₁ and z₂ sit together

We have;

Where the three siblings don't sit together given as $X^c$ ∩ $Y^c$ ∩ $Z^c$

By set theory, we have;

$\left | X^c \cap Y^c \cap Z^c \right |$ = $\left | X^c \cup Y^c \cup Z^c \right |$ = $\left | U \right | - \left | X \cup Y \cup Z \right |$

$\left | U \right | - \left | X \cup Y \cup Z \right |$ = $\left | U \right | - \left (\left | X \right | + \left | Y\right | + \left | Z\right | - \left | X \cap Y\right | - \left | X \cap Z\right | - \left | Y\cap Z\right | + \left | X \cap Y \cap Z\right | \right)$

Therefore;

$\left | X^c \cap Y^c \cap Z^c \right | = \left | U \right | - \left (\left | X \right | + \left | Y\right | + \left | Z\right | - \left | X \cap Y\right | - \left | X \cap Z\right | - \left | Y\cap Z\right | + \left | X \cap Y \cap Z\right | \right)$

Where;

$\left | U\right |$ = The number of ways the 3 pairs of siblings can sit on the 7 chairs = 7!

$\left | X\right |$ = The number of ways x₁ and x₂ can sit together on the 7 chairs = 2 × 6!

$\left | Y\right |$ = The number of ways y₁ and y₂ can sit together on the 7 chairs = 2 × 6!

$\left | Z\right |$ = The number of ways z₁ and z₂ can sit together on the 7 chairs = 2 × 6!

$\left | X \cap Y\right |$ = The number of ways x₁ and x₂ and y₁ and y₂ can sit together on the 7 chairs = 2 × 2 × 5!

$\left | X \cap Z\right |$ = The number of ways x₁ and x₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 5!

$\left | Y \cap Z\right |$ = The number of ways y₁ and y₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 5!

$\left | X \cap Y \cap Z\right |$ = The number of ways x₁ and x₂, y₁ and y₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 2 × 4!

Therefore, we get;

$\left | X^c \cap Y^c \cap Z^c \right |$ = 7! - (2×6! + 2×6! + 2×6! - 2 × 2 × 5! - 2 × 2 × 5! - 2 × 2 × 5! + 2 × 2 × 2 × 4!)

$\left | X^c \cap Y^c \cap Z^c \right |$ = 5,040 - 3072 = 1,968

The number of ways where the three siblings don't sit together given as $\left | X^c \cap Y^c \cap Z^c \right |$ = 1,968

5 0

3 years ago

I need help on this plzzz

antoniya [11.8K]

Answer:

i cant see it

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Travis spies an owl in a tall tree. He estimates the height of the tree to be 85 feet and the angle of elevation to the bird fro

anyanavicka [17]

Answer:

63.44 feet

Step-by-step explanation:

tan(68 degrees)= 85/y

y tan (68 degrees)=85

y=85/tan(68 degrees) ( cross multiply)

y = 34.34 feet

tan (41 degrees)=85/x+y

(x+y)tan(41 degrees)=85

x+y=85/tan 41 degrees

x=85/tan(41 degrees) - y

x= 85/tan (41 degrees)-(34.34) (cross multiply)

Therefore, Travis stepped back 63.44 feet to gain a better view!

7 0

3 years ago