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Tresset [83]
3 years ago
7

Could someone please tell me the answer

Mathematics
1 answer:
Anna007 [38]3 years ago
4 0

Answer: 12.0

Step-by-step explanation: We need to find the length of one of the leg piece, so this is how we will set up our equation (x is the missing side length):

1. x^{2} + 5^{2} = 13^{2} (Simplify the square roots)

2. x^{2} +25 = 169 (Subtract 25 from 169)

3. x^{2} = 144 (Take the square root of both sides)

4. x = 12 or 12.0.

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Solve each system by substitution.<br> y = 2x + 5<br> 3x - y= -6
castortr0y [4]
X=-1 and y=3

To solve this, because y is equal to 2x+5, you can substitute 2x+5 into the bottom equation for y.

You now have 3x - (2x + 5) =-6

This can be solved like a normal equation. Multiply each term in the parenthesis by -1.

3x - 2x - 5 = -6

Subtract 2x from 3x

x - 5 = -6

Add 5 to both sides

x=-1

Now you can solve for y by plugging -1 in for x in either equation. Im going to use the top one.

y= 2(-1)+5
y= -2+5
y=3

You can check your answer by plugging in the values for x and y into both equations and making sure both sides equal each other.
7 0
3 years ago
The polygons are regular polygons. Find the area of the shaded region.
Serga [27]
Polygon area = [circumradius^2 * # sides * sine (360/# sides)] / 2
small hexagon area =[5^2 * 6 * sine (60)] / 2
small hexagon area =[25 * 6 * 0.86603] / 2
<span><span>small hexagon area = 64.95225 </span> sq feet
</span>
large hexagon area =[10^2 * 6 * sine (60)] / 2
large hexagon area =[600 * 0.86603] / 2large hexagon area = <span> <span> <span> 259.809 </span></span></span>sq feet
Area of shaded region = large hexagon area -small hexagon areaArea of shaded region = <span> <span> 259.809</span> -</span>64.95225 <span>Area of shaded region = 194.85675 </span><span>sq feet
</span>
Source:http://www.1728.org/polygon.htm

6 0
3 years ago
Show how you solve the following two equations: 6x = 33 2/3x=8/15
LuckyWell [14K]
6x = 33 x= 33/6. x= 5 3/6 = 5 1/3


2/3x=8/15= x= 8/15÷2/3 x= 8/15 x 3/2 = 24/30 = 4/5
4 0
3 years ago
Read 2 more answers
A semicircle with diameter PQ sits on an isosceles triangle PQR to form a region shaped like a two-dimensional ice- cream cone,
Ksivusya [100]

The limit of a(x)/b(x) as x approaches 0 is gotten as;

lim a(x)/b(x); x→0 = 0

  • The image showing the semi circle and isosceles triangle is missing and so i have attached it.

  • Formula for <em>area</em> of a semi circle is;

a(x) = A = ¹/₂πr²

  • <em>Area</em> of triangle is;

b(x) = A =  ¹/₂ × base × height

  • From the image, ∠PQR = θ

Thus; a(x) = ¹/₂π(10 sin (θ/2))²

a(x) =  ¹/₂π(100 sin² (θ/2))

Similarly;

b(x) =  ¹/₂(20 sin (θ/2)) × (10 cos (θ/2))

b(x) = 100 sin (θ/2) cos (θ/2)

  • Thus;

a(x)/b(x) = ¹/₂π(100 sin² (θ/2)) ÷ 100 sin (θ/2) cos (θ/2)

100 will cancel out and the sin (θ/2) at the denominator will be eradicated

upon further simplification to give;

a(x)/b(x) = ¹/₂π(sin (θ/2)) ÷ cos (θ/2)

In trigonometry, we know that; tan θ = sin θ/cos θ

Thus;

a(x)/b(x) = ¹/₂π(sin (θ/2)) ÷ cos (θ/2) =  ¹/₂π(tan (θ/2))

We want to find the limit as θ approaches <em>zero.</em>

Thus;

lim θ → 0 =  ¹/₂π(tan (0/2))

⇒ ¹/₂π tan 0

⇒ 0

Read more at; https://brainly.in/question/9387458

3 0
2 years ago
Analyze the diagram below and answer the question that follows.
Rama09 [41]

Answer:

Option A is the correct option

BA with one sided arrow mark represent a ray

6 0
3 years ago
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