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astraxan [27]
3 years ago
13

Solve the right triangle. Round your answers to the nearest tenth.

Mathematics
1 answer:
lesya692 [45]3 years ago
8 0

Answer:

sin(41) = 14/c

c = 14/sin(41)

c ≈ 21.34

c ≈ 21.3

a² + 14² = 21.3²

a² + 196 = 453.69

a² = 257.69

a ≈ 16.1

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Find the area of a circle circumscribing an equilateral triangle of side 15cm.<br> Take pi = 3.14.
miskamm [114]

Answer:

236 cm²

Step-by-step explanation:

Height of an equilateral triangle (h) = √3 /2 (l)

l = side of the equilateral triangle.

h = √3 /2 (15)

In an equilateral triangle the orthocenter, centroid, circumcenter and incenter are in the same spot

The center of the circle is the centroid and height match  with the median. The radius of the circumcircle is equal to two thirds the height.

Formula for the Radius of the circumcircle = 2/3 h

= 2/3 x √3 /2 (15)

= 5 √3 cm     (=radius)

Area of the circle = πr^

= 3.14 x( 5 √3 ) ^

=3.14 x(25*3)

=3.14 x 75

=235.5

=236 cm²

8 0
3 years ago
Solve each equation for 37 points
NARA [144]
The answer to A is the pic below

6 0
3 years ago
Read 2 more answers
I need help on number 35
scoray [572]

The symbol they are using is that of a ray so, order matters.

Ray BC is telling you the ray starts at point B like a line segment BUT continues on towards and through point C rather than stopping at it.

Ray CB is telling you just the opposite; the ray starts at point C, continues on towards and through point B.

8 0
3 years ago
Solve 2cos^2x+5cosx+2=0
Nuetrik [128]
2cos^2x+5cosx+2=0\\&#10;x\in\\&#10;\Delta=b^2-4ac\\&#10;\Delta=9\\&#10;\sqrt{\Delta}=3\\&#10;cosx=1\\&#10;or\\&#10;cosx=-6\\&#10;-6\notin\\

Then using trygonometric table or a graph we read that:
cosx=1\Leftrightarrow(x)=0

Period of cosinus function = 2π
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6 0
3 years ago
Find the surface area of the following figure
Julli [10]

Answer:

<h2>510.4 ft²</h2>

Step-by-step explanation:

We have:

two trapezoids with bases 15ft and 7ft and height 5ft.

four rectangles 5ft × 11ft, 15ft × 11ft, 9.4ft × 11ft and 7ft × 11ft.

The formula of an area of a trapezoid:

A_t=\dfrac{b_1+b_2}{2}\cdot h

b₁, b₂ - bases

h - height

Substitute:

A_t=\dfrac{15+7}{2}\cdot5=\dfrac{22}{2}\cdot5=(11)(5)=55\ ft^2

The formula of an area of a rectangle:

A_r=lw

l - length

w - width

The dimensions of rectangle l × w

Subtitute:

A_1=(5)(11)=55\ ft^2\\\\A_2=(15)(11)=165\ ft^2\\\\A_3=(9.4)(11)=103.4\ ft^2\\\\A_4=(7)(11)=77\ ft^2

The surface area of the figure:

[tex]S.A.=2A_t+A_1+A_2+A_3+A_4\\\\S.A.=2(55)+55+165+103.4+77=510.4\ ft^2[/text]

3 0
3 years ago
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