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Kaylis [27]
2 years ago
7

Evaluate 3^−1⋅(4⋅6)⋅2^−3

Mathematics
1 answer:
charle [14.2K]2 years ago
4 0

Answer:

3^{-1}*(4*6)*2^{-3}=1

Step-by-step explanation:

3^{-1}*(4*6)*2^{-3}

\frac{1}{3}*24*\frac{1}{8}

\frac{1}{3}*3

1

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Find the area of a triangle whose sides are 5 m, 6 m and 9m,( use root 2 =-1.41 m)​
dlinn [17]

The area of the triangle is 14.1 square meters

<h3>How to determine the triangle area?</h3>

The side lengths are given as:

5m, 6m and 9m

Calculate the semi-perimeter (s) using

s = (5 + 6 + 9)/2

Evaluate

s = 10

The area is then calculated as:

Area = \sqrt{s(s -a)(s-b)(s-c)

This gives

Area = \sqrt{10 * (10 -5)(10-6)(10-9)

Evaluate the products

Area = \sqrt{200

Evaluate the exponent

Area = 14.1

Hence, the area of the triangle is 14.1 square meters

Read more about areas at:

brainly.com/question/27683633

#SPJ1

4 0
2 years ago
What is the answer for ( c )
Whitepunk [10]

Answer: 37

Step-by-step explanation:

7 0
2 years ago
What is the value of x?
timofeeve [1]

9514 1404 393

Answer:

  x = 3

Step-by-step explanation:

It is convenient to remember the ratios of side lengths of these "special triangles."

The side ratios of ΔABC are 1 : 1 : √2, so BC = AC/√2 = 6.

The side ratios of ΔBCD are 1 : √3 : 2, so BD = BC/2 = 6/2 = 3.

The value of x is 3.

6 0
3 years ago
What is quotient of a number 21
stira [4]
A quotient of a number 21 is 3 and 63
Hope this helps!
8 0
3 years ago
Needdd hellpppppssssssss
Ratling [72]

Answer:

Choice number one:

\displaystyle \frac{5}{10}\cdot \frac{4}{9}.

Step-by-step explanation:

  • Let A be the event that the number on the first card is even.
  • Let B be the event that the number on the second card is even.

The question is asking for the possibility that event A and B happen at the same time. However, whether A occurs or not will influence the probability of B. In other words, A and B are not independent. The probability that both A and B occur needs to be found as the product of

  • the probability that event A occurs, and
  • the probability that event B occurs given that event A occurs.

5 out of the ten numbers are even. The probability that event A occurs is:

\displaystyle P(A) = \frac{5}{10}.

In case A occurs, there will only be four cards with even numbers out of the nine cards that are still in the bag. The conditional probability of getting a second card with an even number on it, given that the first card is even, will be:

\displaystyle P(B|A) = \frac{4}{9}.

The probability that both A and B occurs will be:

\displaystyle P(A \cap B) = P(B\cap A) =  P(A) \cdot P(B|A) = \frac{5}{10}\cdot \frac{4}{9}.

6 0
2 years ago
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