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2 years ago

15

Why do people need pie in there life

1 answer:

tatuchka [14]2 years ago

4 0

Answer:

We dont need pie. Pie is a option. A thing we want

Step-by-step explanation:

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On a coordinate plane, kite W X Y Z is shown. Point W is at (negative 3, 3), point X is at (2, 3), point Y is at (4, negative 4)

xz_007 [3.2K]

Answer:

$P = 10 + 2\sqrt{53}$ units

Step-by-step explanation:

Given

Shape: Kite WXYZ

W (-3, 3), X (2, 3),

Y (4, -4), Z (-3, -2)

Required

Determine perimeter of the kite

First, we need to determine lengths of sides WX, XY, YZ and ZW using distance formula;

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

For WX:

$(x_1, y_1)\ (x_2,y_2) = (-3, 3),\ (2, 3)$

$WX = \sqrt{(-3 - 2)^2 + (3 - 3)^2}$

$WX = \sqrt{(-5)^2 + (0)^2}$

$WX = \sqrt{25}$

$WX = 5$

For XY:

$(x_1, y_1)\ (x_2,y_2) = (2, 3)\ (4,-4)$

$XY = \sqrt{(2 - 4)^2 + (3 - (-4))^2}$

$XY = \sqrt{-2^2 + (3 +4)^2}$

$XY = \sqrt{-2^2 + 7^2}$

$XY = \sqrt{4 + 49}$

$XY = \sqrt{53}$

For YZ:

$(x_1, y_1)\ (x_2,y_2) = (4,-4)\ (-3, -2)$

$YZ = \sqrt{(4 - (-3))^2 + (-4 - (-2))^2}$

$YZ = \sqrt{(4 +3)^2 + (-4 +2)^2}$

$YZ = \sqrt{7^2 + (-2)^2}$

$YZ = \sqrt{49 + 4}$

$YZ = \sqrt{53}$

For ZW:

$(x_1, y_1)\ (x_2,y_2) = (-3, -2)\ (-3, 3)$

$ZW = \sqrt{(-3 - (-3))^2 + (-2 - 3)^2}$

$ZW = \sqrt{(-3 +3)^2 + (-2 - 3)^2}$

$ZW = \sqrt{0^2 + (-5)^2}$

$ZW = \sqrt{0 + 25}$

$ZW = \sqrt{25}$

$ZW = 5$

The Perimeter (P) is as follows:

$P = WX + XY + YZ + ZW$

$P = 5 + \sqrt{53} + \sqrt{53} + 5$

$P = 5 + 5 + \sqrt{53} + \sqrt{53}$

$P = 10 + 2\sqrt{53}$ units

6 0

3 years ago

From the observation deck of a skyscraper, Harper measures a 45^{\circ} ∘ angle of depression to a ship in the harbor below. If

Evgesh-ka [11]

The horizontal distance between the base of the skyscraper to the ship is 1145 feet.

Data;

Angle = 45 degree
Observation deck (adjacent) = 1145 ft
horizontal distance (opposite) = x

<h3>Trigonometric Ratios</h3>

Using trigonometric ratios, we have the value of adjacent and angle, we can easily use tangent of the angle to find the horizontal distance

$tan\theta = \frac{opposite}{adjacent}$

Substitute the values into the equation and solve

$tan45 = \frac{x}{1145}\\ x = 1145tan45\\x = 1145ft$

The horizontal distance between the base of the skyscraper to the ship is 1145 feet.

Learn more on trigonometric ratio here;

brainly.com/question/4326804

5 0

1 year ago

I need help please ?!!!!

Alecsey [184]

Answer:

89

Step-by-step explanation:

= √ 64 +9²

=8 +81

=89

7 0

3 years ago

Read 2 more answers

1+2<br>Please help<br>filler filler filler

Nadya [2.5K]

Answer:

1+2=3 is answer

Step-by-step explanation:

so what help u need?

6 0

2 years ago

The sun shines at a 30° angle to the ground. To the nearest inch, how long is the shadow cast by a 72-inch tall fence post?

Mars2501 [29]

Answer:

The shadow length is 125 inches.

Step-by-step explanation:

The triangle that the sun's rays and the fence post form is shown in the figure.

From trigonometry we know that $tan(30^o)$ is the ratio of the opposite <em>(72-in fence)</em> to the adjacent side $L$ of the triangle <em>(the shadow length): </em>

<em /> $tan(30^o) = \dfrac{72 in}{L}$ <em />

since

<em /> $tan(30^o) = \dfrac{\sqrt{3} }{3}$ <em />

we have

<em /> $\dfrac{\sqrt{3} }{3} = \dfrac{72 in}{L}$ <em />

solving for $L$ we get:

$L = 72\sqrt{3}\\\\\boxed{L = 124.71\: in.}$

Hence, the length of the shadow to the nearest inch is 125 inches.

7 0

2 years ago