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elena-14-01-66 [18.8K]
3 years ago
15

Question 1 options:

Mathematics
1 answer:
NeTakaya3 years ago
3 0

Answer:

x = 18 units.

Area of triangle = 156 units^2.

Step-by-step explanation:

Area of the rectangle =  4(2x + 3) and the area of the triangle = 1/2*6*(3x - 2).

These are equal, so:

4(2x + 3) = 1/2*6*(3x - 2)

8x + 12 = 9x - 6

12 + 6 = 9x - 8x

x = 18.

Area of triangle =  1/2 * 6 * (3(18) - 2)

= 3 * 52

= 156 units^2.

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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
3 years ago
On April 1, 1986, Casey deposited $1150 into a savings account paying 9.6% interest, compounded quarterly. If he hasn't made any
dalvyx [7]
According to the 72 rule
72/rate=time
72÷9.6=7.5 years

Another way to solve by using the main equation
2300=1150(1+0.096/4)^4t
Solve for t
t=((log(2,300÷1,150)÷log(1+(0.096÷4))÷4))=7.31years

Hope it helps :-)
7 0
3 years ago
Read 2 more answers
HELP!!! I HAVE TO DO THIS!! WILL GIVE
sertanlavr [38]

Answer:

1. 9.576

2. 17.442

3. 45.29

4. 79.294

5. 30.736

6. 30.8

7. 31.5

8. 25.85

9. 53.29

10. 49.184

11. 15.18

12. 15.573

13. 17.075

14. 74.06

15. 9.01

Step-by-step explanation:

Use a multiplication method to find the answers. I would put it up here but i dont know how.

3 0
3 years ago
43 = 20 + z/3 please help
BabaBlast [244]

Answer:

z = 69

Step-by-step explanation:

43 - 20 = z/ 3

23 = \frac{z}{3}

cross multiply

z = 23 x 3

z =69

3 0
3 years ago
50 pts!!! Find the area of the shaded regions. Give your answer as a completely simplified exact value in terms of pi (no approx
charle [14.2K]

Answer:

  • 40π

Step-by-step explanation:

<u>Area of circle:</u>

  • A = πr²

<u>Area of α degree sector:</u>

  • S = πr²*α/360

<u>The shaded area is:</u>

  • S = π*10²×(72*2)/360 = 100π×144/360 = 40π
4 0
3 years ago
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