The unit circle with center at the origin is a relation but not a function Find the two functions which are semicircles of the unit circle and determine their domains and ranges?​

Answer:

[tex]y=-3x+8[/tex]

Step-by-step explanation:

Hi there!

What we need to know:

1) Determine the slope (m)

[tex]y = 4 - 3x\\y = -3x+4[/tex]

Given this equation, we can identify the slope to be -3 since it's in the place of m in [tex]y=mx+b[/tex].

Because parallel lines have the same slope, -3 is therefore the slope of the line we're currently solving for. Plug this into [tex]y=mx+b[/tex]:

[tex]y=-3x+b[/tex]

2) Determine the y-intercept (b)

[tex]y=-3x+b[/tex]

Plug in the given point (1,5) and solve for b:

[tex]5=-3(1)+b\\5=-3+b\\8=b[/tex]

Therefore, the y-intercept is 8. Plug this back into [tex]y=-3x+b[/tex]:

[tex]y=-3x+8[/tex]

I hope this helps!

Answer:

the image/question isnt there

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

a^6b^4

Step-by-step explanation:

(a³b²)²

a^6b^4

9514 1404 393

Answer:

  a⁶b⁴

Step-by-step explanation:

The relevant rules of exponents are ...

  (ab)^c = (a^c)(b^c)

  (a^b)^c = a^(bc)

__

These let us simplify the expression as follows:

  [tex](a^3b^2)^2=(a^3)^2(b^2)^2=a^{3\cdot2}b^{2\cdot2}=\boxed{a^6b^4}[/tex]

_____

Additional comment

It might be helpful to remember that an exponent signifies repeated multiplication.

  a·a·a = a³ . . . . . . 'a' is a factor 3 times, so the exponent is 3.

Similarly, ...

  (a³)² = (a³)·(a³) = (a·a·a)·(a·a·a) = a⁶

Answer:

Explaination down below

Step-by-step explanation:

Before you start this question, make sure you know this rule. When multiplying fractions, Multiply the numerator by the other numerator and multiply the denominator by the other one.

[tex]\frac{1}{5}[/tex] x [tex]\frac{1}{2}[/tex]. 1*1=1, so one is the numerator. 2*5= 10 so 10 is the denominator.

The final answer is [tex]\frac{1}{10}[/tex].

You could also look at it another way. What is one half of one-fifth?

That is another way to look at it.

If this helped, please mark me as brainliest. Thank you! ;)

Step-by-step explanation:

Is this the full question

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))

or

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

(c) square root

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

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

and

[tex]\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)[/tex]

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

[tex]\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}[/tex]

[tex]\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}[/tex]

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

[tex]\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}[/tex]

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

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

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

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

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

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

So the two square roots of z are

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

and

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

Answer:

[tex]\displaystyle \text{a. }8+6i\\\\\text{b. }\sqrt{10}\\\\\text{c. }\\\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}}+i\sqrt{\frac{\sqrt{10}-3}{2}},\\-\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}}-i\sqrt{\frac{\sqrt{10}-3}{2}}\\\\\\\text{d. }\\\text{Exact: }z=\sqrt{10}\left(\cos\left(\arctan\left(\frac{1}{3}\right)\right), i\sin\left(\arctan\left(\frac{1}{3}\right)\right)\right),\\\text{Approximated: }z=3.16(\cos(18.4^{\circ}),i\sin(18.4^{\circ}))[/tex]

Step-by-step explanation:

Recall that [tex]i=\sqrt{-1}[/tex]

Part A:

We are just squaring a binomial, so the FOIL method works great. Also, recall that [tex](a+b)^2=a^2+2ab+b^2[/tex].

[tex]z^2=(3+i)^2,\\z^2=3^2+2(3i)+i^2,\\z^2=9+6i-1,\\z^2=\boxed{8+6i}[/tex]

Part B:

The magnitude, or modulus, of some complex number [tex]a+bi[/tex] is given by [tex]\sqrt{a^2+b^2}[/tex].

In [tex]3+i[/tex], assign values:

[tex]|z|=\sqrt{3^2+1^2},\\|z|=\sqrt{9+1},\\|z|=\sqrt{10}[/tex]

Part C:

In Part A, notice that when we square a complex number in the form [tex]a+bi[/tex], our answer is still a complex number in the form

We have:

[tex](c+di)^2=a+bi[/tex]

Expanding, we get:

[tex]c^2+2cdi+(di)^2=a+bi,\\c^2+2cdi+d^2(-1)=a+bi,\\c^2-d^2+2cdi=a+bi[/tex]

This is still in the exact same form as [tex]a+bi[/tex] where:

Thus, we have the following system of equations:

[tex]\begin{cases}c^2-d^2=3,\\2cd=1\end{cases}[/tex]

Divide the second equation by [tex]2d[/tex] to isolate [tex]c[/tex]:

[tex]2cd=1,\\\frac{2cd}{2d}=\frac{1}{2d},\\c=\frac{1}{2d}[/tex]

Substitute this into the first equation:

[tex]\left(\frac{1}{2d}\right)^2-d^2=3,\\\frac{1}{4d^2}-d^2=3,\\1-4d^4=12d^2,\\-4d^4-12d^2+1=0[/tex]

This is a quadratic disguise, let [tex]u=d^2[/tex] and solve like a normal quadratic.

Solving yields:

[tex]d=\pm i \sqrt{\frac{3+\sqrt{10}}{2}},\\d=\pm \sqrt{\frac{{\sqrt{10}-3}}{2}}[/tex]

We stipulate [tex]d\in \mathbb{R}[/tex] and therefore [tex]d=\pm i \sqrt{\frac{3+\sqrt{10}}{2}}[/tex] is extraneous.

Thus, we have the following cases:

[tex]\begin{cases}c^2-\left(\sqrt{\frac{\sqrt{10}-3}{2}}\right)^2=3\\c^2-\left(-\sqrt{\frac{\sqrt{10}-3}{2}}\right)^2=3\end{cases}\\[/tex]

Notice that [tex]\left(\sqrt{\frac{\sqrt{10}-3}{2}}\right)^2=\left(-\sqrt{\frac{\sqrt{10}-3}{2}}\right)^2[/tex]. However, since [tex]2cd=1[/tex], two solutions will be extraneous and we will have only two roots.

Solving, we have:

[tex]\begin{cases}c^2-\left(\sqrt{\frac{\sqrt{10}-3}{2}}\right)^2=3 \\c^2-\left(-\sqrt{\frac{\sqrt{10}-3}{2}}\right)^2=3\end{cases}\\\\c^2-\sqrt{\frac{5}{2}}+\frac{3}{2}=3,\\c=\pm \sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}[/tex]

Given the conditions [tex]c\in \mathbb{R}, d\in \mathbb{R}, 2cd=1[/tex], the solutions to this system of equations are:

[tex]\left(\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}}, \sqrt{\frac{\sqrt{10}-3}{2}}\right),\\\left(-\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}},- \frac{\sqrt{10}-3}{2}}\right)[/tex]

Therefore, the square roots of [tex]z=3+i[/tex] are:

[tex]\sqrt{z}=\boxed{\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}}+i\sqrt{\frac{\sqrt{10}-3}{2}} },\\\sqrt{z}=\boxed{-\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}}-i\sqrt{\frac{\sqrt{10}-3}{2}}}[/tex]

Part D:

The polar form of some complex number [tex]a+bi[/tex] is given by [tex]z=r(\cos \theta+\sin \theta)i[/tex], where [tex]r[/tex] is the modulus of the complex number (as we found in Part B), and [tex]\theta=\arctan(\frac{b}{a})[/tex] (derive from right triangle in a complex plane).

We already found the value of the modulus/magnitude in Part B to be [tex]r=\sqrt{10}[/tex].

The angular polar coordinate [tex]\theta[/tex] is given by [tex]\theta=\arctan(\frac{b}{a})[/tex] and thus is:

[tex]\theta=\arctan(\frac{1}{3}),\\\theta=18.43494882\approx 18.4^{\circ}[/tex]

Therefore, the polar form of [tex]z[/tex] is:

[tex]\displaystyle \text{Exact: }z=\sqrt{10}\left(\cos\left(\arctan\left(\frac{1}{3}\right)\right), i\sin\left(\arctan\left(\frac{1}{3}\right)\right)\right),\\\text{Approximated: }z=3.16(\cos(18.4^{\circ}),i\sin(18.4^{\circ}))[/tex]

Answer:

B

Step-by-step explanation:

i took quiz

n + 1 = 4(n - 8)

n + 1 = 4n - 32

n - 4n = -32 - 1

-3n = -33 / : (-3)

n = 11

Answer:

x=4.25

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

8+8=4x-1

16=4x-1

4x=16+1

4x=17

x= 17/4

I have no idea if i am correct just a guesstimate

Have a good day

Answer:

10

Step-by-step explanation:

Using an exponential function, it is found that the hourly growth rate of the bacteria culture is of 14.9%.

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

An exponential function has the following format:

[tex]A(t) = A(0)(1+r)^t[/tex]

In which:

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

Since it doubles every 5 hours, it means that:

[tex]A(5) = 2A(0)[/tex]

And we use this to find r.

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

[tex]A(t) = A(0)(1+r)^t[/tex]

[tex]2A(0) = A(0)(1+r)^5[/tex]

[tex](1 + r)^5 = 2[/tex]

[tex]\sqrt[5]{(1 + r)^5} = \sqrt[5]{2}[/tex]

[tex]1 + r = 2^{\frac{1}{5}}[/tex]

[tex]1 + r = 1.149[/tex]

[tex]r = 1.149 - 1 = 0.149[/tex]

0.149*100% = 14.9%.

The hourly growth rate of the bacteria culture is of 14.9%.

A similar problem is given at https://brainly.com/question/24218305

Answer:

3/7 or 57.1%

Step-by-step explanation:

If the car wrecks happen 7 times a month and you see 3 wrecks, then you would have seen 3 out of the 7 wrecks. 3/7 or 57.1% of wrecks.

The required probability of observing exactly three accidents is 42.8%.

Give that,
The average number of road accidents that occur on a particular stretch of road during a month is 7. What is the probability of observing exactly three accidents on this stretch of road next month is to be determined.

Probability can be defined as the ratio of favorable outcomes to the total number of events.

here,
Total number of samples = 7
Total number of favorable outcomes = 3
Required probability = 3 / 7 = 42.8%

Thus, the required probability of observing exactly three accidents is 42.8%.

Learn more about probability here:

brainly.com/question/14290572

#SPJ5

Answer:

a) 1.50x + 20 = y

c) $27.5

d) 8 hours

Step-by-step explanation:

a) 1.50 times x the amount of hours. 1.50 per hour. 1 hour would be 1.50 and 2 hours would be 3.00. Then just add 20 to that amount.

b) Don't really have time to graph, but here this is the graph using a graphing calculator. It intercepts at 20 because 20 is b, which is the y intercept.

c) Plug in 5 for x

d) Start with 1.50x + 20 = 32. You need to solve for x. Subtract 20 from both sides and you get 1.50x = 12. Divide both sides by 1.50 and you get x = 8. Since x representshe amount of hours, it would be x.

Answer:

It is correct

Step-by-step explanation:

Has two solutions x = a and x = -a because both numbers are at the distance a from 0. You begin by making it into two separate equations and then solving them separately. An absolute value equation has no solution if the absolute value expression equals a negative number since an absolute value can never be negative.

Answer:

-5/7

Step-by-step explanation:

We need to find g(7) first

All we have to is plug 7 in the place of x

g(7) = 1/7

Now we have f(1/7) = 2x - 1

f(1/7) = 2(1/7) - 1

f(1/7) = 2/7 - 1

Don't forget about common denominators

2/7 - 7/7 = -5/7

Answer:

8i+3j

Step-by-step explanation:

let point P2(0,3)

point P1(-8,0)

vector P1P2= Position vector of P2- position vector of P1

vector= (0,3)-(-8,0)

vector= (8,3)

vector=8i+3j

Answer:

True

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Algebra II

Step-by-step explanation:

Step 1: Define

Identify

Point (5, 6)

Point (-3, 8)

Step 2: Find distance d

Simply plug in the 2 coordinates into the distance formula to find distance d

Answer:

6 in.

Step-by-step explanation:

To find the area of the rectangle, substitute the given values into the formula A = lw. Then, substitute 36 inches squared into the formula A = s2. Find the square root of both sides of the equation to find s.

the correct answer would be 54 degrees. you would subtract 90 and 54 from 360, leaving 108. you would have to divide by 2 since we have 1 and 2 angles, that leaves 54. hope this helps!

Answer:

last option 180 degrees

Step-by-step explanation:

let ∠1 and angle ∠2 be 2x as both are of same angles

angle sum proprty = 360

90 + 54+ ∠1 +∠2 = 360

144 + 2x = 360

2x = 360 - 144

2x = 216

x = 216 ÷ 2

x = 108

∠2 = 108 degrees

Answer:

1.01

Step-by-step explanation:

1^2= 1

0.1^2= 0.01

1+0.01=1.01

Given : The length of a rectangular garden is 4 feet longer than the width. If the perimeter is 192 feet, what is the area of the garden ?

Solution :

Let us assume the breadth be x

The length is 4 ft longer than the Breadth

So, the length be x + 4

Perimeter = 192

❍ Perimeter = 2(Length + Breadth)

Length : x + 4 = 46 + 4 = 50

Breadth : x = 46

Step-by-step explanation:

14 - 2(x + 8) = 5x - (3x - 34)

14 -2x -16 = 5x -3x+34

-2x -2 = 2x+34

-2x-2x = 34+2

-4x = 36

x = 36/(-4)

x = -9

Answer:

2/7x + 8/7

Step-by-step explanation:

f(x) = -4 + 7/2x

y = -4+7/2x

Exchange x and y

x = -4 +7/2y

Solve for y

Add 4 to each side

x+4 = -4+4 +7/2y

x+4 = 7/2y

Multiply each side by 2/7

2/7(x+4) = 2/7 * 7/2y

2/7(x+4) = y

2/7x + 8/7

Answer: New price =  rs. 1710

Step-by-step explanation:

Given information

Original price (market price) = rs.1800

Discount rate = 5%

Given expression deducted from the question

New price = Original price × (1 - discount rate)

Substitute values into the expression

New price = 1800 × (1 - 5%)

Simplify parentheses

New price = 1800 × 0.95

Simplify by multiplication

New price =[tex]\boxed{rs.1710}[/tex]

Hope this helps!! :)

Please let me know if you have any questions

Answer:

+400

Step-by-step explanation:

Answer:

amount in 30 yrs is: $1787.0977

interest is for 20 yrs: $1387.0977

Step-by-step explanation:

Given data

principal amount = $400

rate = 5 % = 0.05

time period = 20 years

(Sorry if I'm wrong :( )

3²: It is an example of numeric expression

is a mathematical sentence that encompasses, power, root, multiplication, division, addition and subtraction.

In this question - the following example 3² was given. This example can be classified as a numerical expression, because a power is a multiplication of equal factors.

So, this example is a numeric expression.

Answer:

6 * 100 = 600

14 * 10  = 140

12 * 1    = 12

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

752

Answer:

6,9

Step-by-step explanation:

The two multiples of 3

which are more than 5 and less than 10 are

6 and 9

b. The pattern grow by adding 1 tile above the tile and adding 1 tile at the right of the tile.

c. In figure 0, there will be 1 tile. We know this because in each successive figures a tile is added at the above and a tile is added to the right, so ineach preceeding figure the same is reduced. In figure 1, there are e tiles, so in figure 0, there will be 3-2 = 1 tile.