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cricket20 [7]
3 years ago
9

Find the area of following rhombuses. Round your answers to the nearest tenth if necessary.

Mathematics
1 answer:
polet [3.4K]3 years ago
7 0

Answer:

Area =55.4ft^2

Step-by-step explanation:

Given

The attached rhombus

Required

The area

First, calculate the length of half the vertical diagonal (x).

Length x is represented as the adjacent to 60 degrees

So, we have:

\tan(60) = \frac{4\sqrt 3}{x}

Solve for x

x = \frac{4\sqrt 3}{\tan(60)}

\tan(60) = \sqrt 3

So:

x = \frac{4\sqrt 3}{\sqrt 3}

x = 4

At this point, we have established that the rhombus is made up 4 triangles of the following dimensions

Base = 4\sqrt 3

Height = 4

So, the area of the rhombus is 4 times the area of 1 triangle

Area = 4 * \frac{1}{2} * Base * Height

Area = 4 * \frac{1}{2} * 4\sqrt 3 * 4

Area =2 * 4\sqrt 3 * 4

Area =55.4ft^2

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The graph shows two lines, A and B.
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Answer:

second option: (4, 4) is the solution to both lines A and B.

Step-by-step explanation:

You know that the equation of line A is:

y=-3x+15

and the equation of line B is:

y=\frac{1}{2}x+2

The point in which the line A intersects with the line B is the solution of the sytstem of equations.

You can observe in given graph that the point of intersection of Line A and Line B is: (4,4)

Therefore (4, 4) is the solution to both lines A and B.

8 0
3 years ago
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given that sine theta = 0, and cosine theta is less than zero, find the exact values of the other two trig functions
vekshin1

Answer:

θ = π + periods of 2π

Sin (π + 2π) = 0

Cos (π + 2π) = -1

Tan (π + 2π) = 0

Step-by-step explanation:

Sin (θ)=0 implies that θ only can be 0 or π plus periods of 2π:

θ = 0+2π

θ = π+2π

For Cos(θ) the values only can be:

Cos (0+2π) = 1 and

Cos (π+2π) = -1

from this, only Cos (π+2π) < 0

So θ only can be θ=π+2π

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3 years ago
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Show work and explain with formulas:
Marina86 [1]

17 Answer:  205

<u>Step-by-step explanation:</u>

\{1\dfrac{2}{3}+1\dfrac{5}{6}+2+...+8\dfrac{1}{3}\}\implies a_1=1\dfrac{2}{3},\ d=\dfrac{1}{6}\\\\\\a_n=a_1+d(n-1)\qquad solve\ for\ n\\\\8\dfrac{1}{3}=1\dfrac{2}{3}+\dfrac{1}{6}(n-1)\\\\\\\dfrac{25}{3}=\dfrac{5}{3}+\dfrac{1}{6}n-\dfrac{1}{6}\\\\\\\dfrac{50}{6}=\dfrac{10}{6}+\dfrac{1}{6}n-\dfrac{1}{6}\\\\\\\dfrac{41}{6}=\dfrac{1}{6}n\\\\\\\dfrac{41}{6}\cdot 6=n\\\\41=n

\text{Now use the sum formula:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\\S_{41}=\dfrac{1\dfrac{2}{3}+8\dfrac{1}{3}}{2}\cdot 41\\\\\\.\quad =\dfrac{10}{2}\cdot 41\\\\\\.\quad =5\cdot 41\\\\.\quad =\large\boxed{205}

18 Answer:  1968

<u>Step-by-step explanation:</u>

a_1=-6,\ d=2,\ n=48, \quad \text{solve for }a_{48}\\\\a_{n}=a_1+d(n-1)\\\\a_{48}=-6+2(48-1)\\\\.\quad =-6+2(47)\\\\.\quad =-6+94\\\\.\quad =88

\text{Now use the sum formula:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\\S_{48}=\dfrac{-6+88}{2}\cdot 48\\\\\\.\quad =\dfrac{82}{2}\cdot 48\\\\\\.\quad =41\cdot 48\\\\.\quad =\large\boxed{1968}

19 Answer:  -116

<u>Step-by-step explanation:</u>

\{3, -2, -7, ...\}\\a_1=3,\ d=-5,\ n=8, \quad \text{solve for }a_{8}\\\\a_{n}=a_1+d(n-1)\\\\a_{8}=3-5(8-1)\\\\.\quad =3-5(7)\\\\.\quad =3-35\\\\.\quad =-32\\\\\text{Now use the sum formula:}\\\\S_8=\dfrac{a_1+a_8}{2}\cdot 8\\\\\\S_{8}=\dfrac{3-32}{2}\cdot 8\\\\\\.\quad =\dfrac{-29}{2}\cdot 8\\\\\\.\quad =-29\cdot 4\\\\.\quad =\large\boxed{-116}

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