For -180°<θ<0 , which of the primary trigonometric functions may have positive values?

D

A

B

C

for more explanation please don't hesitate to just respond

Answer:

y = 3

Step-by-step explanation:

6sin(30) = 3

Answer:

6

Step-by-step explanation:

First, we can expand the function to get its expanded form and to figure out what degree it is. For a polynomial function with one variable, the degree is the largest exponent value (once fully expanded/simplified) of the entire function that is connected to a variable. For example, x²+1 has a degree of 2, as 2 is the largest exponent value connected to a variable. Similarly, x³+2^5 has a degree of 2 as 5 is not an exponent value connected to a variable.

Expanding, we get

(x³-3x+1)²  = (x³-3x+1)(x³-3x+1)

= x^6 - 3x^4 +x³ - 3x^4 +9x²-3x + x³-3x+1

= x^6 - 6x^4 + 2x³ +9x²-6x + 1

In this function, the largest exponential value connected to the variable, x, is 6. Therefore, this is to the 6th degree. The fundamental theorem of algebra states that a polynomial of degree n has n roots, and as this is of degree 6, this has 6 roots

Answer:

it 94x26.2 i think it right if not sorry :/

Step-by-step explanation:

Answer:

See Below.

Step-by-step explanation:

We are given the equation:

[tex]\displaystyle y^2 = 1 + \sin x[/tex]

And we want to prove that:

[tex]\displaystyle 2y\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right) ^2 + y^2 = 1[/tex]

Find the first derivative by taking the derivative of both sides with respect to x:

[tex]\displaystyle 2y \frac{dy}{dx} = \cos x[/tex]

Divide both sides by 2y:

Find the second derivative using the quotient rule:

[tex]\displaystyle \begin{aligned} \frac{d^2y}{dx^2} &= \frac{(\cos x)'(2y) - (\cos x)(2y)'}{(2y)^2}\\ \\ &= \frac{-2y\sin x-2\cos x \dfrac{dy}{dx}}{4y^2} \\ \\ &= -\frac{y\sin x + \cos x\left(\dfrac{\cos x}{2y}\right)}{2y^2} \\ \\ &= -\frac{2y^2\sin x+\cos ^2 x}{4y^3}\end{aligned}[/tex]

Substitute:

[tex]\displaystyle 2y\left(-\frac{2y^2\sin x+\cos ^2 x}{4y^3}\right) + 2\left(\frac{\cos x}{2y}\right)^2 +y^2 = 1[/tex]

Simplify:

[tex]\displaystyle \frac{-2y^2\sin x-\cos ^2x}{2y^2} + \frac{\cos ^2 x}{2y^2} + y^2 = 1[/tex]

Combine fractions:

[tex]\displaystyle \frac{\left(-2y^2\sin x -\cos^2 x\right)+\left(\cos ^2 x\right)}{2y^2} + y^2 = 1[/tex]

Simplify:

[tex]\displaystyle \frac{-2y^2\sin x }{2y^2} + y^2 = 1[/tex]

Cancel:

[tex]\displaystyle -\sin x + y^2 = 1[/tex]

Substitute:

[tex]-\sin x + \left( 1 + \sin x\right) =1[/tex]

Simplify. Hence:

[tex]1\stackrel{\checkmark}{=}1[/tex]

Q.E.D.

there are 12 set of months in a year

Answer:

The new coordinates are [tex]A'(x,y) = (3, 0)[/tex], [tex]B'(x,y) = (6, -3)[/tex] and [tex]C'(x,y) = (2, -3)[/tex].

Step-by-step explanation:

Vectorially speaking, the translation of a point can be defined by the following expression:

[tex]V'(x,y) = V(x,y) + T(x,y)[/tex] (1)

Where:

[tex]V(x,y)[/tex] - Original point.

[tex]V'(x,y)[/tex] - Translated point.

[tex]T(x,y)[/tex] - Translation vector.

If we know that [tex]A(x,y) = (-3,4)[/tex], [tex]B(x,y) = (0,1)[/tex], [tex]C(x,y) = (-4,1)[/tex] and [tex]T(x,y) = (6, -4)[/tex], then the resulting points are:

[tex]A'(x,y) = (-3, 4) + (6, -4)[/tex]

[tex]A'(x,y) = (3, 0)[/tex]

[tex]B'(x,y) = (0,1) + (6, -4)[/tex]

[tex]B'(x,y) = (6, -3)[/tex]

[tex]C'(x,y) = (-4, 1) + (6, -4)[/tex]

[tex]C'(x,y) = (2, -3)[/tex]

The new coordinates are [tex]A'(x,y) = (3, 0)[/tex], [tex]B'(x,y) = (6, -3)[/tex] and [tex]C'(x,y) = (2, -3)[/tex].

Answer:

D

The missing piece (triangle) is facing right side up but the red triangle has its point facing left

TO get it facing up, turn it by 90 degrees clockwise

Answer:  75% - these are books on mathematics.

Step-by-step explanation:

[tex]\dfrac{3}{4} \cdot 64 = 48\\[/tex]

64 - 100%

48  -  x%

[tex]\dfrac{64}{100} =\dfrac{48}{x}\\\\x=\dfrac{48 \cdot 100}{64} \\\\x=75 \%[/tex]

Answer:

5:12

Step-by-step explanation:

125:300 simplified = 5:12

I hope this helps

Answer: a = 25.67

Step-by-step explanation:

minus sign ironically makes it go to the right

because the function crosses the y axis at -13

It is the graph of y = x translated 13 units down is the statement describes a transformation of the graph of y=x.

Graph is a mathematical representation of a network and it describes the relationship between lines and points.

The equation y = x - 13 represents a transformation of the graph of y = x. To find the type of  transformation, we have to compare the two equations and look for changes.

In the equation y = x - 13, we subtract 13 from the value of x.

This means that the graph of y = x is shifted 13 units downwards,

since every point on the graph has 13 subtracted from its y-coordinate.

Hence,  It is the graph of y = x translated 13 units down is the statement describes a transformation of the graph of y=x.

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Answer:

Given E(t)=1100(100-t)+580(100-t)^2

Put t = 99.5, we get

E(99.5)=1100(100-99.5)+580(100-99.5)^2

E(99.5)=1100(0.5)+580(0.5)^2

E(99.5)=1100(0.5)+580(0.25)

E(99.5)=550+145

E(99.5)=695m

Step-by-step explanation:

It can be concluded that -

E(99.5) = 695

E(100) = 0

In mathematics, an expression or mathematical expression is a finite combination of symbols that is well-formed according to rules that depend on the context.

Mathematical symbols can designate numbers (constants), variables, operations, functions, brackets, punctuation, and grouping to help determine order of operations and other aspects of logical syntax.

Given is the function as follows -

E(t) = 1100(100 - t) + 580(100 - t)²

The given function is -

E(t) = 1100(100 - t) + 580(100 - t)²

At → E(99.5)

E(99.5) = 1100(100 - t) + 580(100 - t)²

E(99.5) = 1100(100 - 99.5) + 580(100 - 99.5)²

E(99.5) = 1100(0.5) + 580(0.5)²

E(99.5) = 550 + 145

E(99.5) = 695

At → E(100)

E(100) = 1100(100 - t) + 580(100 - t)²

E(100) = 1100(100 - 100) + 580(100 - 100)²

E(100) = 0

Therefore, it can be concluded that -

E(99.5) = 695

E(100) = 0

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Two workers finished a job in 7.5 days.

How long would it take each worker to do the job by himself if one of the workers needs 8 more days to finish the job than the other worker?

let t = time required by one worker to complete the job alone

then

(t+8) = time required by the other worker (shirker)

let the completed job = 1

A typical shared work equation

7.5%2Ft + 7.5%2F%28%28t%2B8%29%29 = 1

multiply by t(t+8), cancel the denominators, and you have

7.5(t+8) + 7.5t = t(t+8)

7.5t + 60 + 7.5t = t^2 + 8t

15t + 60 = t^2 + 8t

form a quadratic equation on the right

0 = t^2 + 8t - 15t - 60

t^2 - 7t - 60 = 0

Factor easily to

(t-12) (t+5) = 0

the positive solution is all we want here

t = 12 days, the first guy working alone

then

the shirker would struggle thru the job in 20 days.

Answer:7 + 17 = 24÷2 (since there are 2 workers) =12. Also, ½(7) + ½17 = 3.5 + 8.5 = 12. So, we know that the faster worker will take 7 days and the slower worker will take 17 days. Hope this helps! jul15

Step-by-step explanation:

Answer:

60

Step-by-step explanation:

10 x 6 = 60

Hope this helped! :)

More information pleaseeeeeeee

Answer:

jhshejwjabsgsgshshsnsjs

Answer:

Step-by-step explanation:

Put the equation into center-radius form.

x² + y² + 8x - 12y + 24 = 0

x² + y² + 8x - 12y = -24

(x²+8x) + (y²-12y) = -24

(x²+8x+4²) + (y²-12y+6²) = 4²+6²-24

(x+4)² + (y-6)² = 28

Center: (-4,6)

radius: √28

y'' - 6y' + 9y = 0

If y = C₁ exp(3x) + C₂ x exp(3x), then

y' = 3C₁ exp(3x) + C₂ (exp(3x) + 3x exp(3x))

y'' = 9C₁ exp(3x) + C₂ (6 exp(3x) + 9x exp(3x))

Substituting these into the DE gives

(9C₁ exp(3x) + C₂ (6 exp(3x) + 9x exp(3x)))

… … … - 6 (3C₁ exp(3x) + C₂ (exp(3x) + 3x exp(3x)))

… … … + 9 (C₁ exp(3x) + C₂ x exp(3x))

= 9C₁ exp(3x) + 6C₂ exp(3x) + 9C₂ x exp(3x))

… … … - 18C₁ exp(3x) - 6C₂ (exp(3x) - 18x exp(3x))

… … … + 9C₁ exp(3x) + 9C₂ x exp(3x)

= 0

so the provided solution does satisfy the DE.

Answer:

A cable that weighs 6 lb/ft is used to lift 600 lb of coal up a mine shaft 500 ft deep. Find the work done. Show how to approximate the required work by a Riemann sum.

Step-by-step explanation:

Answer:

[tex]\angle P[/tex]

Step-by-step explanation:

Given

[tex]\triangle PRQ = \triangle TSU = 90^o[/tex]

[tex]PQ = 20[/tex]     [tex]QR = 16[/tex]    [tex]PR = 12[/tex]

[tex]ST = 30[/tex]       [tex]TU = 34[/tex]    [tex]SU = 16[/tex]

See attachment

Required

Which sine of angle is equivalent to [tex]\frac{4}{5}[/tex]

Considering [tex]\triangle PQR[/tex]

We have:

[tex]\sin(P) = \frac{QR}{PQ}[/tex] --- i.e. opposite/hypotenuse

So, we have:

[tex]\sin(P) = \frac{16}{20}[/tex]

Divide by 4

[tex]\sin(P) = \frac{4}{5}[/tex]

Hence:

[tex]\angle P[/tex] is correct

Answer:

A or <P

Step-by-step explanation:

on edge 2021

Answer:

f(x)=(x-1)^2+5 with domain x>1 and range y>5 has inverse g(x)=sqrt(x-5)+1 with domain x>5 and range y>1.

Step-by-step explanation:

The function is a parabola when graphed. It is in vertex form f(x)=a(x-h)^2+k where (h,k) is vertex and a tells us if it's reflected or not or if it's stretched. The thing we need to notice is the vertex because if we cut the graph with a vertical line here the curve will be one to one. So the vertex is (1,5). Let's restrict the domain so x >1.

* if x>1, then x-1>0.

* Also since the parabola opens up, then y>5.

So let's solve y=(x-1)^2+5 for x.

Subtract 5 on both sides:

y-5=(x-1)^2

Take square root of both sides:

Plus/minus sqrt(y-5)=x-1

We want x-1>0:

Sqrt(y-5)=x-1

Add 1 on both sides:

Sqrt(y-5)+1=x

Swap x and y:

Sqrt(x-5)+1=y

x>5

y>1

Answer:

Step-by-step explanation:

yeah of course..your answer is true.. it's part d and there is no solution for that system...w and v both of them remove in solution of system and we don't have any unknown variable for solving.

9514 1404 393

Answer:

  D. no solutions

Step-by-step explanation:

The first equation can be simplified to standard form:

  0.5(8w +2v) = 3

  4w +v = 3 . . . . . . . eliminate parentheses

__

The second equation can also be simplified to a comparable standard form:

  8w = 2 -v +4w

  4w +v = 2 . . . . . . add v -4w

__

Comparing these two equations, we find the variable expressions to be the same, but the constants to be different. If any set of variable values were to satisfy one of these equations, it could not satisfy the other equation. There are no solutions to the system.

Answer:

Step-by-step explanation:

the arrows from the picture tells us that TV is parallel to RS

since TS is a transversal that cuts the 2 parallel lines TV and RS than ∠S =x

(alternate interior angles)

sum of angles in a Δ is 180° so x+x+2x = 180°, 4x =180°, x= 45°

2x = 45*2 = 90°

Answer:

In picture.

Step-by-step explanation:

To do this answer, you need to count the boxes up to the mirror line. This will give us the exact place to draw the triangle.

The picture below is the answer.

Answer:

20'×15 in 54 inches

Step-by-step explanation:

The Best as a pool should be rectangular in shape and 54inches deep for safety of life's

The volume of the rectangular pool that is 20' x 15' in 54 inches deep is largest.

The volume of the cylinder is the product of the height, pie, and square of the radius.

The volume of the cylinder = [tex]\pi r^{2}[/tex]h

The volume of the cylindrical pool that has a 3.3 m radius and is 1.8 m deep is;

=  [tex]\pi r^{2}[/tex]h

[tex]= 3.14 (3.3)^2 (1.8)\\\\= 61.55 m^3[/tex]

The volume of the rectangular pool that is 20' x 15' in 54 inches deep ;

V = 20 x 15 x 54

V = 16,200 cubic meter.

The volume of the rectangular pool that is 20' x 15' in 54 inches deep is largest.

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Answer:

35

Step-by-step explanation:

y = 35x+23 is in the form

y = mx+b  where m is the slope and b is the y intercept

The slope can also be called the rate of change

35 is the slope

Answer:

12 cm²

Step-by-step explanation:

area = L *B

a = 6 * 2

a = 12cm²