Find the cube root of 4 - 4√3i
that graphs in the second
quadrant.
[?] (cos[]° + i sin[__ _]°)
Use degree measure.
Enter

Answer:

1/2

Step-by-step explanation:

13 total fish

remove one fish

12 total fish

6 total clown fish

6/12 simplified is 1/2

By the quadratic formula, we find that the two zeroes of the quadratic function y = x² - 20 · x + 32 are x₁ = 10 + 2√17 and x₂ = 10 - 2√17, respectively.

A value of x is a zero of a polynomial if and only if [tex]\sum\limits_{i=0}^{n} c_{i}\cdot x^{i} = 0[/tex], the quadratic formula for second order polynomials of the form a · x² + b · x + c = 0 is presented below:

[tex]x =\frac{-b \pm \sqrt{b^{2}-4\cdot a \cdot c}}{2\cdot a}[/tex]

If we know that a = 1, b = -20 and c = 32, then the roots of the second order polynomial are:

[tex]x = \frac{20 \pm \sqrt{(-20)^{2}-4\cdot (1)\cdot (32)}}{2\cdot (1)}[/tex]

[tex]x = 10 \pm 2\sqrt{17}[/tex]

By the quadratic formula, we find that the two zeroes of the quadratic function y = x² - 20 · x + 32 are x₁ = 10 + 2√17 and x₂ = 10 - 2√17, respectively.

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Answer: C

Step-by-step explanation: I know it trust me

The slope of the linear equation, y = -9x - 8 is determined as: -9.

If the equation of a line is given in a slope-intercept form, y = mx + b, the slope value is represented as m.

Thus, given the equation y = -9x - 8, the slope in the linear equation is -9.

Therefore, slope of y = -9x - 8 is: -9.

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Answer:

[tex]y=-5x-8[/tex]

Step-by-step explanation:

Given the following question:

To calculate the slope intercept of a line/point we need to use the formula for slope intercept, substitute, and solve for b.

Formula for slope intercept:
[tex]y=mx+b[/tex]
[tex]m=-5[/tex]
[tex]y=2[/tex]
[tex]x=-2[/tex]
[tex]2=-5(-2)+b[/tex]

Solve for b:
[tex]2=-5(-2)+b[/tex]
[tex]-5(-2)=10[/tex]
[tex]2=10+b[/tex]
[tex]10-10=0[/tex]
[tex]2-10=-8[/tex]
[tex]-8=b[/tex]
[tex]b=-8[/tex]

Substitute for the following:
[tex]y=mx+b[/tex]
[tex]b=-8[/tex]
[tex]m=-5[/tex]
[tex]y=-5x-8[/tex]

Your answer is "y = -5x - 8."

Hope this helps.

Answer:

y=3

Step-by-step explanation:

y=mx+b where m = slope and b=y-intercept
slope: 0, y-intercept: 3

y=(0)x + 3

y=3

Inequalities help us to compare two unequal expressions.  The inequality that represents losing more than 500 points is x<-500.

Inequalities help us to compare two unequal expressions. Also, it helps us to compare the non-equal expressions so that an equation can be formed.

It is mostly denoted by the symbol <, >, ≤, and ≥.

Given In a computer game positive numbers represent earning points and negative numbers represent losing points. Therefore, the loss of more than 500 points is given by the inequality,

x< -500

Hence, the inequality that represents losing more than 500 points is x<-500.

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71% subtracted from 100% is 29%

Add or subtract the given percent from 100%.

71%

The percentage (p) is given as:

p = 71%

When subtracted from 100%, we have:

100% - p

Substitute p = 71%

100% - 71%

71 subtracted from 100 is 29.

So, we have:

100% - 71% = 29%

Hence, 71% subtracted from 100% is 29%

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Answer:

-13/20

Step-by-step explanation:

-0.65=-65/100

=13/20

Using translation concepts, it is found that graph 3 represents the function [tex]f(x) = \sqrt{x} + 1[/tex].

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.

In this problem, the function is:

[tex]f(x) = \sqrt{x} + 1[/tex]

Which is a shift up of one unit of [tex]\sqrt{x}[/tex]. The square root function has y-intercept at (0,0), hence the translated function will have y-intercept at (0,1). Additionally, [tex]\sqrt{4} = 2[/tex], since f(4) = 3, which means that graph 3 is represents the function.

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Using it's concept, it is found that the probability that the first student chosen is a senior and the second student chosen is a sophomore is given by:

[tex]p = \frac{11}{320}[/tex]

A probability is given by the number of desired outcomes divided by the number of total outcomes.

Researching this problem on the internet, we have that there are:

31 juniors, 10 sophomores, 17 juniors, 22 seniors.

Then:

The events are independent, hence the probability of both is given by:

[tex]p = p_1p_2 = \frac{11}{40} \times \frac{1}{8} = \frac{11}{320}[/tex]

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Answer:

Exact Form:−2/3

Decimal Form:−0.666…

Step-by-step explanation:

Answer:

13.3 km/h

Step-by-step explanation:

getting the neutral is adding both the hours and distance, after adding them you can then dicixe the distance by the time getting your average speed (the upstream is what balances the downstream)

The range of salary is (a) the range of salaries is from $38,500 to $39,800.

The given parameters are:

Average salary = 39,800

Vary = Less than $1300

Let the salary be x.

So, we have:

x = 39,800

When the salary varies, the equation becomes

x = 39800 - 1300

Evaluate

x = 38500

This means that the salary is from 38500 to 39800

Hence, the range of salary is (a) the range of salaries is from $38,500 to $39,800.

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Answer:

y<_0

Step-by-step explanation:

The last answer.
Hope this helps! Have a great day :)

The graph that shows the solution set for the system of inequalities is graph C.

Here we have the system of linear inequalities:

[tex]y < (-1/3)*x + 1\\\\y \geq 2x- 1[/tex]

For the first one, we can see a line with a negative slope, and y is strictly smaller than that. So we will have a dashed line that goes down, and we need to shade the region below that line.

For the second inequality, we can see that y can be equal to or larger than the line, so this time we will use a solid line (because the points on the line are solutions). and we will shade above that line.

Also notice that this time the slope is positive, so this line goes upwards.

With that general description, we conclude that the correct option will be graph C.

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First product

Second and final product

Answer:

Given polynomials

The product of the given polynomials:

[tex]\begin{aligned}(2ab+b)(a^2-b^2) & = 2ab(a^2-b^2)+b(a^2-b^2)\\& = 2a^3b-2ab^3+a^2b-b^3\end{aligned}[/tex]

The product of the given polynomials multiplied by [tex](2a+b)[/tex]:

[tex]\begin{aligned}(2a+b)(2a^3b-2ab^3+a^2b-b^3) & = 2a(2a^3b-2ab^3+a^2b-b^3)+b(2a^3b-2ab^3+a^2b-b^3)\\& = 4a^4b-4a^2b^3+2a^3b-2ab^3+2a^3b^2-2ab^4+a^2b^2-b^4\\& = 4a^4b+2a^3b^2-4a^2b^3-2ab^4+2a^3b+a^2b^2-2ab^3-b^4\end{aligned}[/tex]

Therefore, the resulting polynomial has 8 terms.

Let A(t) denote the amount of salt (in ounces, oz) in the tank at time t (in minutes, min).

Salt flows in at a rate of

[tex]\dfrac{dA}{dt}_{\rm in} = \left(17 (1 + 15 \sin(t)) \dfrac{\rm oz}{\rm gal}\right) \left(8\dfrac{\rm gal}{\rm min}\right) = 136 (1 + 15 \sin(t)) \dfrac{\rm oz}{\min}[/tex]

and flows out at a rate of

[tex]\dfrac{dA}{dt}_{\rm out} = \left(\dfrac{A(t) \, \mathrm{oz}}{180 \,\mathrm{gal} + \left(8\frac{\rm gal}{\rm min} - 8\frac{\rm gal}{\rm min}\right) (t \, \mathrm{min})}\right) \left(8 \dfrac{\rm gal}{\rm min}\right) = \dfrac{A(t)}{180} \dfrac{\rm oz}{\rm min}[/tex]

so that the net rate of change in the amount of salt in the tank is given by the linear differential equation

[tex]\dfrac{dA}{dt} = \dfrac{dA}{dt}_{\rm in} - \dfrac{dA}{dt}_{\rm out} \iff \dfrac{dA}{dt} + \dfrac{A(t)}{180} = 136 (1 + 15 \sin(t))[/tex]

Multiply both sides by the integrating factor, [tex]e^{t/180}[/tex], and rewrite the left side as the derivative of a product.

[tex]e^{t/180} \dfrac{dA}{dt} + e^{t/180} \dfrac{A(t)}{180} = 136 e^{t/180} (1 + 15 \sin(t))[/tex]

[tex]\dfrac d{dt}\left[e^{t/180} A(t)\right] = 136 e^{t/180} (1 + 15 \sin(t))[/tex]

Integrate both sides with respect to t (integrate the right side by parts):

[tex]\displaystyle \int \frac d{dt}\left[e^{t/180} A(t)\right] \, dt = 136 \int e^{t/180} (1 + 15 \sin(t)) \, dt[/tex]

[tex]\displaystyle e^{t/180} A(t) = \left(24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t)\right) e^{t/180} + C[/tex]

Solve for A(t) :

[tex]\displaystyle A(t) = 24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t) + C e^{-t/180}[/tex]

The tank starts with A(0) = 15 oz of salt; use this to solve for the constant C.

[tex]\displaystyle 15 = 24,480 - \frac{66,096,000}{32,401} + C \implies C = -\dfrac{726,594,465}{32,401}[/tex]

So,

[tex]\displaystyle A(t) = 24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t) - \frac{726,594,465}{32,401} e^{-t/180}[/tex]

Recall the angle-sum identity for cosine:

[tex]R \cos(x-\theta) = R \cos(\theta) \cos(x) + R \sin(\theta) \sin(x)[/tex]

so that we can condense the trigonometric terms in A(t). Solve for R and θ :

[tex]R \cos(\theta) = -\dfrac{66,096,000}{32,401}[/tex]

[tex]R \sin(\theta) = \dfrac{367,200}{32,401}[/tex]

Recall the Pythagorean identity and definition of tangent,

[tex]\cos^2(x) + \sin^2(x) = 1[/tex]

[tex]\tan(x) = \dfrac{\sin(x)}{\cos(x)}[/tex]

Then

[tex]R^2 \cos^2(\theta) + R^2 \sin^2(\theta) = R^2 = \dfrac{134,835,840,000}{32,401} \implies R = \dfrac{367,200}{\sqrt{32,401}}[/tex]

and

[tex]\dfrac{R \sin(\theta)}{R \cos(\theta)} = \tan(\theta) = -\dfrac{367,200}{66,096,000} = -\dfrac1{180} \\\\ \implies \theta = -\tan^{-1}\left(\dfrac1{180}\right) = -\cot^{-1}(180)[/tex]

so we can rewrite A(t) as

[tex]\displaystyle A(t) = 24,480 + \frac{367,200}{\sqrt{32,401}} \cos\left(t + \cot^{-1}(180)\right) - \frac{726,594,465}{32,401} e^{-t/180}[/tex]

As t goes to infinity, the exponential term will converge to zero. Meanwhile the cosine term will oscillate between -1 and 1, so that A(t) will oscillate about the constant level of 24,480 oz between the extreme values of

[tex]24,480 - \dfrac{267,200}{\sqrt{32,401}} \approx 22,995.6 \,\mathrm{oz}[/tex]

and

[tex]24,480 + \dfrac{267,200}{\sqrt{32,401}} \approx 25,964.4 \,\mathrm{oz}[/tex]

which is to say, with amplitude

[tex]2 \times \dfrac{267,200}{\sqrt{32,401}} \approx \mathbf{2,968.84 \,oz}[/tex]

Answer:

[tex]r = \dfrac{p}{2 + \pi}[/tex]

Step-by-step explanation:

[tex]~~~~~p = 2r +\pi r\\\\\implies p = r(2 + \pi)\\\\\implies r = \dfrac{p}{2 + \pi}[/tex]

Answer:

Step-by-step explanation:

-2/3 * (-1/4) =

2/3 * 1/4 =

2/12 =

1/6

Answer:

The "what" is 100

Step-by-step explanation:

Rewrite this as 37.6y = 3760 and set out to determine y:

y = 3760/37.6  = 100

The "what" is 100

Answer:

[tex]\left(12x^5-3x^1+2x-5\right)+\left(8x^1-3x^3+4x+1\right)[/tex]

=> Expand

[tex]12x^5-3x^1+2x-5+8x^1-3x^3+4x+1[/tex]

=> simplify and group like terms

[tex]12x^5-3x^3+5x^1+6x-4[/tex]

=> simplify further

[tex]12x^5-3x^3+11x-4[/tex]

There are 43 small dogs signed up in the competition

Represent the small dogs with x and the large dogs with y.

So, we have the following equations

x + y = 50

x = 36 + y

Make y the subject in x = 36 + y

y = x - 36

Substitute y = x - 36 in x + y = 50

x + x - 36 = 50

Evaluate the like terms

2x = 86

Divide both sides by 2

x = 43

Hence, there are 43 small dogs in the competition

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Answer:

B.

Step-by-step explanation:

Because the point on the top on the original would still be on top so it is 100% B

Hope This Helped

Answer:

Vertex(1,7)

Vertex form: [tex]f(x)=-2\left(x-1\right)^2+7[/tex]

Step-by-step explanation:

Vertex form equations for x is [tex]v=-\frac{b}{2a}[/tex]

which the vertex here is

[tex]-\frac{4}{2\left(-2\right)}[/tex]

->

x = 1, plugin to find y, y = 7

To calculate the cost of the fence, we have to find the length of the semi-circle. The cost of fencing the garden would be $328.52 approximately

To find the length of the fence, we have to calculate the circumference of the semi-circle and divide it by 2

[tex]C = 2\pi r\\l = \frac{c}{2} \\l = \pi r[/tex]

Data;

Let's substitute the values and calculate the length of the fence.

[tex]l = \pi r\\l = 3.14 * 13.5\\l = 42.39ft[/tex]

The length of the fence is equal to 42.39ft

The cost of fencing would be

[tex]42.39 * 7.75 = $328.5225[/tex]

The cost of fencing the garden would be $328.52 approximately

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Hello, the answer of each option is given below.

Good luck! These are four-operation-equations. We should calculate by our hand or try to use calculator to get the result. Anyway, If you have any question(s), then feel free to ask in comments!

The probability that a player will be asked a math question and then a music question is 0.03125

The size of the sections are given as:

Music = 0.5

Sport = 0.5

Others = 1

So, the total size is:

Total = 0.5 + 0.5 + 1 + 1 + 1

Evaluate

Total = 4

The probability of asking a math question is:

P(Math) = 1/4 = 0.25

The probability of asking a music question is:

P(Music) = 0.5/4 = 0.125

The required probability is:

P(Math and Music) = P(Math) * P(Music)

This gives

P(Math and Music) = 0.25 * 0.125

Evaluate

P(Math and Music) = 0.03125

Hence, the probability that a player will be asked a math question and then a music question is 0.03125

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