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gulaghasi [49]
1 year ago
14

Type the correct answer in each box. Round your answers to the nearest integer.

Mathematics
2 answers:
QveST [7]1 year ago
5 0
Did you ever find the awnser

Over [174]1 year ago
3 0

Answer:

(1) 92.92 unit  (2) 500 unit²

Step-by-step explanation:

(1) Perimeter of the given hexagon ABCDEF will be

   = Sum of length of all sides

   = AB + BC + CD + DE + EF + FA

   =10+(\sqrt{10^{2}+10^{2}})+\sqrt{10^{2}+10^{2}}+10+\sqrt{20^{2}+10^{2}}+\sqrt{20^{2} +10^{2} }

   = 20+10\sqrt{2} +10\sqrt{2}+10+10\sqrt{(2+1)}+10\sqrt{(2+1)}

   = (30+20\sqrt{2}+20\sqrt{3})

   = 30 + 28.28 + 34.64

   = 92.92 unit

(2) Area of hexagon = Area of ΔAEF + Area of rectangle ABDE + Area of ΔBCD

   = \frac{1}{2} (20)(20)+(20)(10)+\frac{1}{2}(20)(10)

   = 200 + 200 + 100

   = 500 unit²

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bearhunter [10]
I = prt

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t = 3

I = (7430)(0.05)(3)
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4 0
1 year ago
Read 2 more answers
Chen is bringing fruit and veggies to serve at an afternoon meeting. He spends a total of $28.70 on 5 pints of cut veggies and 7
deff fn [24]
The answer would be $1.75 (Rounded off to the nearest cent).

Substitute the given cost of each pint of fruit to the equation. Transpose and simplify, to get the price of a pint of veggies.
Given: <span>5v + 7f = 28.70; f = $2.85
Solution: </span><span>5v + 7(2.85) = 28.70
                </span><span>5v + 19.95 = 28.70
                5v = 8.75
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6 0
2 years ago
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Use the position function s(t) = −16t² + 400, which gives the height (in feet) of an object that has fallen for t seconds from a
skad [1K]

Answer:

160m/s

Step-by-step explanation:

The object can hit the ground when t = a; meaning that s(a) = s(t) = 0

So, 0 = -16a² + 400

16a² = 400

a² = 25

a = √25

a = 5 (positive 5 only because that's the only physical solution)

The instantaneous velocity is

v(a) = lim(t->a) [s(t) - s(a)]/[t-a)

Where s(t) = -16t² + 400

and s(a) = -16a² + 400

v(a) = Lim(t->a) [-16t² + 400 + 16a² - 400]/(t-a)

v(a) = Lim(t->a) (-16t² + 16a²)/(t-a)

v(a) = lim (t->a) -16(t² - a²)(t-a)

v(a) = -16lim t->a (t²-a²)(t-a)

v(a) = -16lim t->a (t-a)(t+a)/(t-a)

v(a) = -16lim t->a (t+a)

But a = t

So, we have

v(a) = -16lim t->a 2a

v(a) = -32lim t->a (a)

v(a) = -32 * 5

v(a) = -160

Velocity = 160m/s

7 0
1 year ago
Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:
professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

a) We need to find a relation R reflexive and transitive that contain the relation R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

  • 1R4 and 4R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 4R1 and 1R2, then 4 must be related with 2.

Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

  • since 1R2, then 2 must be related with 1.
  • since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

  • 1 must be related with 1,
  • 2 must be related with 2,
  • 3 must be related with 3,
  • 4 must be related with 4

For be symmetric

  • since 1R2, 2 must be related with 1,
  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

5 0
1 year ago
A trampoline salesman makes $25,000 annually plus 6% commission on his total sales. If he sold $40,000 worth of trampolines this
Katarina [22]

Multiply sales by 6%

40,000 x 0.06 = 2,400

Add that to his base pay:

2400 + 25000 = $27,400 total pay.

5 0
1 year ago
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