2. Find the value of x in the triangle. Round your answer to the nearest tenth of a degree.

Drag the tiles to the correct boxes to complete the pairs. Match the polar equations to their rectangular forms.

LekaFEV [45]

The correct answer for the polar coordinates will be C, A, E, B, D

<h3>What is a polar coordinate system?</h3>

The polar coordinate system is a two-dimensional coordinate system in which a distance from a reference point and an angle from a reference direction identify each point on a plane. The pole is the reference point, and the polar axis is the ray from the pole in the reference direction.

The conversion relations between rectangular and polar coordinates can be used to find the polar coordinate equivalents. Those relations are ...

x = r·cos(θ)

y = r·sin(θ)

x² +y² = r²

x = 2

Using the expression for x, this is ...

r·cos(θ) = 2

r = 2/cos(θ) . . . . divide by cos(θ)

r = 2·sec(θ) . . . . use the trig identity

x²+y²=36

From above, this becomes ...

r² = 36

r = 6 . . . . . . take the square root

x²+y²=2y

From above, this becomes ...

r² = 2·r·sin(θ)

r = 2·sin(θ) . . . . . . divide by r

x=√3y

Using the above relations, this is ...

r·cos(θ) = √3·r·sin(θ)

1/√3 = sin(θ)/cos(θ) . . . . . . divide by √3·r·cos(θ)

tan(θ) = 1/√3 ⇒ θ = arctan(1/√3)

θ = π/6

x=y

From above, this is ...

r·cos(θ) = r·sin(θ)

1 = sin(θ)/cos(θ) = tan(θ) . . . . divide by r·cos(θ)

θ = arctan(1)

θ = π/4

However, the tangent function is periodic with period π, so θ = arctan(k)+π is also equivalent to the rectangular equation. It is the opposite ray, so forms the complete line when joined with the first ray.

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4 0

1 year ago

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The endpoints of CD are C(3,−5) and D(7, 9).<br> The coordinates of the midpoint M of CD are?

Nutka1998 [239]

Answer:

M(5,2)

Step-by-step explanation:

We use the midpoint rule:

$( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} )$

We just have to find the arithmetic mean of the x and y-coordinates.

The endpoints of CD are given as C(3,−5) and D(7, 9).

We substitute these values to get:

$( \frac{7 + 3}{2}, \frac{9 + - 5}{2} )$

$( \frac{10}{2}, \frac{4}{2} )$

$(5, 2)$

Therefore the coordinates of the midpoint M of CD are (5,2)

3 0

3 years ago

Determine the area of the figure

aliya0001 [1]

The area of the trapezoid is x+11 cm.

Given a trapezoid has one side of (x+6) cm, the other side is 5 cm, and the height is 2 cm.

A two-dimensional quadrilateral consisting of the sum of non-adjacent parallel sides and a suitable non-parallel side is denoted as a trapezoid.

Now we will use the formula to find the area of a trapezoid

Area = 1/2 × (base 1 + base 2) × height

Here base 1 = (x+6) cm, base 2 = 5cm and height = 2cm

Substituting the values into the formula, we get

Area = 1/2 × (x+6+5) × 2

Area = 1/2 × (x+11) × 2

Area = x+11

Thus, the area of the given figure when one side is x+6, the other side is 5 cm and the height is 2 cm is x+11 cm.

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8 0

2 years ago

The managers of 21 supermarkets counted the number of cars in their parking lots on the same day. The results are shown in the l

mixer [17]

The IQR is 42.5

Step-by-step explanation:

Interquartile range is the difference of third and first quartile.

First of all we have to find the median for that purpose the data has to be arranged in ascending order. The data is already in ascending order.

As the number of values are odd

n=21

The median will be: $(\frac{n+1}{2}) th\ term$

Putting n=21

$(\frac{21+1}{2})th\ term\\=(\frac{22}{2})th\ term\\= 11th\ term$

The 11th term is 133

So median = 133

Now the data is divided into two halves

One is: 98, 100, 101, 102, 108, 109, 111,118, 129, 132

2nd is: 135, 135, 145, 146, 146, 156, 170 176, 180, 180

Q1 will be the median of first half and Q3 will be the median of 2nd half.

As now the halves contain even number of values, the medians will be the average of middle two values

<u>For First Half:</u>

98, 100, 101, 102, <u>108, 109</u>, 111,118, 129, 132

$Q_1 = \frac{108+109}{2}\\Q_1 = \frac{217}{2}\\Q_1 = 108.5$

<u>For Second Half:</u>

135, 135, 145, 146, 146, 156, 170 176, 180, 180

$Q_2 = \frac{146+156}{2}\\Q_2 = \frac{302}{2}\\Q_2 = 151$

Now

<u>Interquartile Range:</u>

$IQR = Q_3-Q_1\\= 151-108.5\\=42.5$

Hence,

The IQR is 42.5

Keywords: Median, IQR

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7 0

3 years ago

Evaluate 10 to the 7th

Olenka [21]

Ya came to the right place! Its 10,000,000 fam

5 0

3 years ago

Read 2 more answers