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umka21 [38]
3 years ago
10

Can anyone help me? I'm having lot's of trouble.

Mathematics
1 answer:
Kaylis [27]3 years ago
8 0
I don't quite understand yet, but I think you have to count the squares on the line between the letters. Does this help?
You might be interested in
Solve for x. What is the value of<br><br> Pls help me
vfiekz [6]

Answer

x = 21

HOS = 68

Explanation

The arrows beside x point to 12 and 6, which is a straight angle. Straight angles always add up to 180 degrees.

So that means 3x+5 plus 5x+7 must equal 180 degrees.

3x+5+5x+7 = 180

Combine like terms.

8x+12 = 180

Subtract 12 from both sides.

8x = 168

x = 21

But BE CAREFUL.

The problem is asking for the angle HOS.

To do this, just substitute what you got for x into the expression.

3(21)+5 = 68

The value of HOS is 68 degrees.

8 0
3 years ago
Question 1 of 10
nordsb [41]

Answer:

B.

Step-by-step explanation: I took the test.

3 0
2 years ago
suppose a marble is chosen randomly from a bag containing 3 blue, 2 red, and 5 green marbles. Determine each theoretical probabi
givi [52]

Answer:

3/10 + 2/10 +5/10

3/10+ 1/5 + 1/2

4 0
3 years ago
What is an equivalent expression of 12x + 10 + 4y?
saveliy_v [14]

Answer:

Step-by-step explanation:

2(6x+5+2y)

7 0
2 years ago
Read 2 more answers
In how many ways can you place 50 identical balls in six distinct urns such that each urn contains an odd number of balls
mezya [45]

Answer:

80730

Step-by-step explanation:

The answer is c(27,5)=27*26*25*24*23/5!. Here is why.

Let us solve a simpler problem. In how many ways 25 identical balls can be placed in six distinct urns such that no urn is empty?

25 balls in a row make 24 places for five dividers, so the answer is c(24,5).

2. Next problem. In how many ways 50 identical balls can be placed in six distinct urns such that no urn is empty and each urn contains even number of balls.

We group 50 balls in 25 pairs, so the answer is the same as in the previous problem which is

c(48/2,5)=c(24,5).

Let us denote the content of each urn as

u(1),u(2),…u(6). Note that each u(i) is even. Denote u(i) as u(i)=2k(i) where k is a natural number. We have

2k(1)+2k(2)+2k(3)+2k(4)+2k(5)+2k(6)=50.

3. Now let us proceed to the original problem where odd number of balls in each urn is required.

We have

2k(1)-1+2k(2)-1+2k(3)-1+2k(4)-1+2k(5)-1+2k(6)-1=50 or

2k(1)+2k(2)+2k(3)+2k(4)+2k(5)+2k(6)=56.

So the original problem reduces to previous problem where even number of balls in each urn is required but the total number of balls is 56 instead of 50.

Its answer is c(54/2,5)=c(27,5).

8 0
2 years ago
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