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Mumz [18]
2 years ago
6

Can someone please solve

Mathematics
1 answer:
BartSMP [9]2 years ago
8 0

Answer: \sqrt{7}

Step-by-step explanation:

Using the Pythagorean theorem,

3^2 + b^2 = 4^2\\\\9+b^2 = 16\\\\b^2 = 7\\\\b=\boxed{\sqrt{7}}

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Multiple choice!! There's only one correct answer.
GalinKa [24]

The solution is where the red line crosses the blue line, which on the X axis is between the 1 and the 2, so about 1.5.

Convert the fractions to decimal and see which one is  close to 1.5:

13/8 = 1.625

25/16 = 1.5625

7/4 = 1.75

27/16 = 1.6875

The closest one to 1.5 is 25/16

5 0
3 years ago
Alberts cabbage patch has 12 rows of cabbage. in each row, there are 15 heads of cabbage. how many cabbage does Albert have in a
Sedbober [7]
180 cabbages is how many he has
7 0
2 years ago
Read 2 more answers
A membership committee of three is formed from four eligible members. Let the eligible members be represented by A, B, C, and D.
yKpoI14uk [10]

The correct statement is:

  • There are four ways to choose the committee.
  • There are three ways to form the committee if person D must be on it.
  • If persons B and C must be on the committee, there are two ways to form the committee.

It is given that:

A membership committee of three is formed from four eligible members.

1) There are four ways to choose the committee.

This statement is true.

Since we have to choose 3 members out of the 4 members so we can use the method of combination.

2) There are three ways to form the committee if person D must be on it.

This statement is also true.

Since D has to be in the committee, this means we have to choose 2 more people out of the three people to form the committee.

3) If seven members are eligible next year, then there will be fewer combinations.

This statement is wrong.

Since we have to choose 3 members out of 7 members so the number of possible combinations will be:

There are 35 combinations possible.

4) If persons B and C must be on the committee, there are two ways to form the committee.

If B and C have to be in the committee then we have to choose just one person out of the two people left.

Hence, the statement is true.

5) If persons A and C must be on the committee, then there is only one way to form the committee.

If A and C have to be on the committee then as in the last option we have to choose any one of the two-person left.

So possible number of ways is 2.

Hence, the statement is false.

Learn more about committee here: brainly.com/question/425830

#SPJ4

<em>Your question is incomplete. Please read below to find the missing content.</em>

<em />

A membership committee of three is formed from four eligible members. Let the eligible members be represented by A, B, C, and D. The possible outcomes include S = {ABC, ABD, ACD, BCD}.

Which statements about the situation are true? Check all that apply.

There are four ways to choose the committee.

There are three ways to form the committee if person D must be on it.

If seven members are eligible next year, then there will be fewer combinations.

If persons B and C must be on the committee, there are two ways to form the committee.

If persons A and C must be on the committee, then there is only one way to form the committee.

8 0
1 year ago
Solve the following equation 1÷x=2​
Troyanec [42]

Answer:

Your answer would be x = 1/2

Step-by-step explanation:

1/x = 2

You would divide x on both sides...then they would cancel each other out.

But now you're left with...

1 = x•2

So, re-order the terms so constants are on the left

1 = 2x

Divide 2 on both sides...

Now you're left with

1/2=x or more practically it would be x = 1/2

Hope this helps!

5 0
2 years ago
I don't understand how to do this,<br> 6x + 6/6 &gt; 1/3x + 2
-BARSIC- [3]

Answer:

X> 3/17

Step-by-step explanation:

6x+ 6/6 > 1/3x + 2

5 0
3 years ago
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