Use the angle difference identity to sovle 5cos2(theta) =1 for 0 degrees is less than or equal to theta is less than 360 degrees

Answers

Answer 1
Answer: use cos (2∅)= 2cos² ∅ -1

5(2cos²∅ -1 )=1
10cos²∅ -5 = 1
10cos²∅ = 6
cos²∅ = 0.6
cos∅ = ± √(0.6)
∅=  39.2,180 - 39.2, 180 + 39.2 , 360 - 39.2
∅ = 39.2 , 140.8 , 219.2 , 320.8 

Related Questions

A contestant in a game show has​ 9,000 points. The contestant answers the next question incorrectly and loses 750 points. Does the word​ "lose" suggest a positive integer or a negative​ integer? What integer represents a loss of 750​ points?
Let L1 be the line passing through the points Q1=(−1, 5, 2) and Q2=(−2, 7, 3) and let L2 be the line passing through the point P1=(−8, 4, 9) with direction vector →d=[−6, 3, 6]T. Determine whether L1 and L2 intersect. If so, find the point of intersection Q.
Given ABCDEFGH, what rigid motion,or composition of rigid motions maps ABCD to EFGH? I need help immediately!!!
Write an equation that is perpendicular to the line y = 2x + 13?
Ans is 16/9 . But I want the method​

___% of 750 = 37.5

20

15

10

7

5

Answers

Answer:

let y% of 750 = 37.5

y = 3750/750

= 5

I think is 5 not sure

2. The restaurant bill you and your friend received was not itemized. You ordered 2 drink refills and the egg breakfast, for a total of $8.40. Your friend ordered 3 refills and the egg breakfast for $9.35. (a) Write a system of equations for the situation. Use r for the amount charged for a refill and b for the amount charged for the egg breakfast.
(b) Graph the equations in the system.
(c) Use your graph to estimate how much each item cost.

Answers

( a )   The system of equations:
2 r + b = 8.40
3 r + b = 9.35
( b ) Graph is in the attachment.
( c )  Each item costs:
 b = $6.50,  r = $0.95
We can prove it:  2 * 0.95 + 6.50 = 8.40
3* 0.95 + 6.50 = 9.35

Answer: 1 drink refill is $0.95 and the egg breakfast is $6.50

2 r + b = 8.40

3 r + b = 9.35

$9.35-$8.40 =$0.95 then take .95x2 to get 1.90 and 8.40-1.90 to get the egg breakfast or $6.50

Write an equation of the parabola in vertex form.

Answers

Answer:

y= -\frac19(x-3)^2+1

Step-by-step explanation:

the x-3 is to shift a normal parabola 3 to the right

the +1 is to lift the top up by one

the -1/9 is to compress and swapt it around

Answer:

y =-  (1)/(9)(x - 3)² + 1

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Here (h, k) = (3, 1) , thus

y = a(x - 3)² + 1

To find a substitute another point on the graph into the equation

Substitute (1, (5)/(9) ) into the equation

(5)/(9) = a(1 - 3)² + 1 ( subtract 1 from both sides )

- (4)/(9) = 4a ( divide both sides by 4 )

a = - (1)/(9)

y = - (1)/(9) (x - 3)² + 1 ← equation of parabola in vertex form

Find an equation in standard form for the ellipse with the vertical major axis of length 18 and minor axis of length 16.

Answers

Answer:

x²/64 + y²/81 = 1

Step-by-step explanation:

Standard form of an equation for the ellipse is (x^(2) )/(a^(2))+(y^(2) )/(b^(2) )=1

Here b is the length of vertical major axis = 9

and minor axis of length a = 8

Therefore the equation of the ellipse will be

(x^(2) )/(8^(2) )+(y^(2) )/(9^(2) ) =1

(x^(2) )/(64)+(y^(2) )/(81)=1

So the answer is x²/64 + y²/81 = 1

Let us assume then that the center is the origin.  If the major axis is 18, then a = 9 and a^2=81.  If the minor axis is 16, then b = 8 and b^2=64.  Now you can write the equation.  Remember that this ellipse is vertical and so a^2 goes under y^2

And entryway floor is square with an area of 81 feet^2 which side length of a square rug would fit in the entryway with a border of 1 foot of floor around each edge? O 11 ft O 9ft O 8ft O7ft

Answers

Answer:

D.) 7 ft

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

I think it’s A sorry if I’m wrong

Solve.

√(x-1)+3=x

Answers

√(x-1)+3=x;\ Domain:x-1\geq0\ \wedge\ x\geq3\to x\geq1\ \wedge\ x\geq3\nD:x\in[3;\ \infty)\n\n√(x-1)=x-3\ \ \ \ |square\ both\ sides\n\nx-1=(x-3)^2\ \ \ \ |use\ of\ the\ formula:(a-b)^2=a^2-2ab+b^2\n\nx-1=x^2-6x+9\n\nx^2-6x+9=x-1\n\nx^2-6x-x+9+1=0\nx^2-7x+10=0\ \ \ \ |use\ quadratic\ formula

a=1;\ b=-7;\ c=10\n\n\Delta=b^2-4ac\to\Delta=(-7)^2-4\cdot1\cdot10=49-40=9\n\nx_1=(-b-\sqrt\Delta)/(2a);\ x_2=(-b+\sqrt\Delta)/(2a)\n\n\sqrt\Delta=\sqrt9=3\n\nx_1=(7-3)/(2\cdot1)=(4)/(2)=2\notin D\n\nx_2=(7+3)/(2\cdot1)=(10)/(2)=5\in D\n\nSolution:x=5
√(x-1)+3=x\n √(x-1)=x-3\n D:x-1\geq0 \wedge x-3\geq0\n D:x\geq1 \wedge x\geq3\n D:x\geq3\n x-1=(x-3)^2\n x-1=x^2-6x+9\n x^2-7x+10=0\n x^2-2x-5x+10=0\n x(x-2)-5(x-2)=0\n (x-5)(x-2)=0\n x=5 \vee x=2\n 2\not \in D \Rightarrow x=5