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MAXImum [283]
3 years ago
14

Create three problems that require you to solve each using each of the trig ratios (one problem where you will use sine, one pro

blem where you will use tangent and one problem where you will use cosine) and then solve each problem, show all your work.

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
8 0
See the attached image for the drawings of the problems. The figures are not to scale. The decimal values in each figure are approximate to one decimal place.

----------------------------------------------

Focus on Figure 1

sin(angle) = opposite/hypotenuse
sin(A) = BC/AC
sin(36) = 13.2/x
x*sin(36) = 13.2
x = 13.2/sin(36)
x = 13.2/0.58778525229248 <<-- make sure calc is in degree mode
x = 22.4571813404936
x = 22.5
This value is approximate (rounded to one decimal place)

----------------------------------------------

Move onto Figure 2

cos(angle) = adjacent/hypotenuse
cos(D) = DE/FD
cos(50) = y/57.4
57.4*cos(50) = y
y = 57.4*cos(50)
y = 57.4*0.64278760968653
y = 36.8960087960069
y = 36.9
This value is approximate (rounded to one decimal place)

----------------------------------------------

Finally, move onto Figure 3

tan(angle) = opposite/adjacent
tan(G) = JH/GH
tan(18) = z/10
10*tan(18) = z
z = 10*tan(18)
z = 10*0.3249196962329
z = 3.249196962329
z = 3.2
This value is approximate (rounded to one decimal place)

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Estimate the perimeter of the figure to the nearest tenth.
pantera1 [17]

Answer:

<em>The Perimeter of the Figure to the nearest tenth is 18.7 units</em>

Step-by-step explanation:

<em>Please note I have attached an edited version of your sketch to aid my solution. Now this question can be solved in multiple ways. Here, we shall see one of them. </em>

<em />

Looking at the original sketch, we can see that the figure is actually a combination of a Triangle (Figure 1 in my sketch) and a rectangle (Figure 2 in my sketch). So we can simply find the sides of a Triangle and the sides of a Rectangle and add them.

<u>Perimeter on Figure 1:</u>

The Perimeter of a Triangle is given by the Sum of the three sides as:

A_{T}=a+b+c

<u>Perimeter on Figure 2:</u>

The Perimeter of the Rectangle is given by:

A_{R}=2(w+l)

where here w is the width and l is the length of the rectangle.

<u>Perimeter on Sketched Figure:</u>

<u><em>The perimeter of the total figure will be two sides of the triangle and the three sides of the rectangle (as the one adjacent between Fig. 1 and 2 can not be taken into account). Thus we need to find 5 different sides and add them together. </em></u>

Now since the figure is on a <em>graph paper</em>, we can read of the size of some sides, thus the left side of the triangle is a=3 units and the base of the triangle is also b=3 units. Now to find the last unknown side we can take Pythagorian theorem, since our triangle is a Right triangle, (i.e. one angle is 90°).

<em>Pythagoras states that the squared of the hypotenuse of a right triangle is equal to the sum of the squares of the other two legs of the triangle (where the hypotenuse side is always across the 90° angle</em>. So here we can say that:

h^2=a^2+b^2\\

where h is the hypotenuse and our unknown side. So plugging in values and solving for  h we have:

h=\sqrt{a^2+b^2}\\ h=\sqrt{3^2+3^2}\\\\h=\sqrt{9+9}\\ h=\sqrt{18}\\ h=3\sqrt{2} units.

So from figure 1 we now have two known sides:

a=3\\h=3\sqrt{2}

Now lets find the other three sides from Figure 2. So here similarly we can read from the graph paper that the base of the rectangle (i.e. the width w) is w=3 and the lengths of the rectangle, are equal and also of the same length as the hypotenuse on the triangle steps. Thus from Figure 2 we have the three remaining sides as:

w=3\\l=h=3\sqrt{2}

So finally the Perimeter of the Sketched Total Figure will be the summation of all 5 sides as:

\Pi_{figure}=a+h+w+l+l\\\Pi_{figure}=3+3\sqrt{2}+3+2(3\sqrt{2} )\\\Pi_{figure}= 6+9\sqrt{2}  \\\Pi_{figure}=18.7 units

(rounded to the nearest tenth)

5 0
4 years ago
Evaluate the following expression. 6^-3
zysi [14]

Answer: 0.00462962962...

Step-by-step explanation:

8 0
3 years ago
Identify the LCM for 6 and 18.<br> 36<br> 108<br> 18<br> 6
oee [108]

Answer: 18

Step-by-step explanation:

8 0
3 years ago
Replace the * with a monomial so that the trinomial may be represented by a square of a binomial:
pogonyaev

Answer:

Below in bold.

Step-by-step explanation:

The identity is of the form

(a + b)^2 = a^2 + 2ab + b^2.

a) Sqrt 49 = 7 and we need + 28 as the middle coefficient . We get this with

2*7 + 2*7 so the first coefficient is 2*2 = 4.

So * = 4a^2.

(2a + 7)^2 = (2a + 7)(2a + 7) = 4x^2 + 14a + 14a + 49.

b) -6 * -6 = 36   -24 =  2*-6 + 2 *-6 so the last term is 4x^2

c) The middle term  must be an 'ab' term.

sqrt 6.25 = 2.5 and sqrt 1/4 = 1/2

So the coefficient of the middle term is 2.5 * 1/2 + 2.5 * 1/2

= 2.5

So the middle term is 2.5ab.

d) The first term will be in b^2.

100 = 10* 10 and we need 2 as a middle term so coefficient of the first term

will be  1/100 or 0.01.  as the 2 comes from 0.1 * 10 + 0.01 * 10 and (0.1)^2 = 0.01

So it is 0.01b^2.

8 0
3 years ago
In the coordinate plane, find the distance between the points using absolute value.
alukav5142 [94]

Answer:

The distance between the two points is 8 units

Step-by-step explanation:

In the given figure

∵ The coordinates of the 2 points are (4, 5) and (4, -3)

∴ x1 = 4 and y1 = 5

∴ x2 = 4 and y2 = -3

∵ Their x-coordinates are equal

→ That means the distance between them is vertical

∴ The length of the vertical distance between two points is

   the difference between the y-coordinates of the points

∵ The distance between them = Iy2 - y1I

∵ y1 = 5 and y2 = -3

∴ The distance between them = I-3 - 5I

∴ The distance between them = I-8I

→ Remember absolute value gives always positive values

∴ The distance between them = 8 units

∴ The distance between the two points is 8 units

8 0
3 years ago
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