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Alex73 [517]
2 years ago
15

Find the measure of Angle A 65°

Mathematics
1 answer:
REY [17]2 years ago
7 0

Answer:

The supplement of 65° is the angle that when added to 65° forms a straight angle (180°).

Step-by-step explanation:

Hope I <u><em>Helped!</em></u> :D

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The product of a particular number and a second value, which is given by the sum of 8 and the original number, is 65. What is th
Marina86 [1]

Given:

The product of a particular number is 65.

Second value is given by the sum of 8 and the original number.

To find:

The unknown number.

Solution:

Let x be the unknown number.

Then, second number = x+8

The product of a particular number is 65.

x\times (x+8)=65

x^2+8x=65

x^2+8x-65=0

Splitting the middle term, we get

x^2+13x-5x-65=0

x(x+13)-5(x+13)=0

(x-5)(x+13)=0

Using zero product property, we get

x-5=0 and x+13=0

x=5 and x=-13

So, the unknown number is either 5 or -13.

Therefore, the correct option is B.

8 0
2 years ago
An urn contains 13 red marbles and 14 blue marbles. 16 marbles are chosen at random. what is the probability that 6 red marbles
11111nata11111 [884]
6/43
which means 
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4 0
3 years ago
Find the volume. Help asap. Please.
alexgriva [62]

Use the Pythagorean theorem to find the diameter:

Diameter = √(19.3^2 - 9.5^2)

Diameter = √(372.49 - 90.25)

Diameter = √282.24

Diameter = 16.8 m

Volume of a cylinder = PI x r^2 x h

r = 1/2 diameter = 16.8 /2 = 8.4

h = 9.5 m

Volume = PI x 8.4^2 x 9.5

= PI x 70.56 x 9.5

= PI x 670.32

In terms of PI volume = 670.32PI

As a decimal:

670.32 x 3.14 = 2104.8048 = 2100m^3 ( rounded to the nearest hundred)

5 0
3 years ago
Read 2 more answers
What is the LCM of 306 and 270
kap26 [50]

Answer:

LCM = 4590

Step-by-step explanation:

1. Find the prime factorization of 306

306 = 2 × 3 × 3 × 17

2. Find the prime factorization of 270

270 = 2 × 3 × 3 × 3 × 5

3. Multiply each factor the greater number of times it occurs in steps i) or ii) above to find the lcm:

LCM = 2 × 3 × 3 × 3 × 5 × 17


6 0
2 years ago
Read 2 more answers
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
2 years ago
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