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Vlad [161]
3 years ago
8

Find two numbers with a sum of 20 and a difference of 14.. . A. 4 and 18. . B. –3 and –17. . C. 2 and 16. . D. 3 and 18

Mathematics
2 answers:
kotegsom [21]3 years ago
6 0
From the given equation 
X+Y=20-->i
X-Y=14--->ii
add eq i and eq ii
2x=34
x=17
putting the value of x=17 in eq i
17+y=20
then y = 3 so the right answer is 
17 and 3
zmey [24]3 years ago
6 0
Hello,

A) 4+18=22  no
B) -3+(-17)=-20 no
C) 2+16=18 no
D) 3+18=21 no

If E is 3 and 17 then YES!!!!!
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Answer:

0.016

Step-by-step explanation:

u divide 0.048 by 30 and the result is the answer

4 0
2 years ago
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How many possible committees of 4 people can be chosen from 15 men and 20 women so that at least two women should be on each com
Mnenie [13.5K]

<span>If there has to be 2 men and 2 women, we know that we must take a group of 2 men out of the group of 15 men and a group of 2 women out of the group of 20 women. Therefore, we have:

(15 choose 2) x (20 choose 2)


(15 choose 2) = 105

(20 choose 2) = 190

190*105 = 19950

Therefore, there are 19950 ways to have a group of 4 with 2 men and 2women.</span>

 

<span>If there has to be 1 man and 3 women, we know that we must take a group of 1 man out of the group of 15 men and a group of 3 women out of the group of 20 women. Therefore, we have:

(15 choose 1) x (20 choose 3)


(15 choose 1) = 15

(20 choose 3) = 1140

15*1140 = 17100

Therefore, there are 17100 ways to have a group of 4 with 3 women and 1 man.</span>

 

<span>We now find the total outcomes of having a group with 4 women.

We know this is the same as saying (20 choose 4) = 4845</span>

Therefore, there are 4845 ways to have a group of 4 with 4 women.

 

We now add the outcomes of 2 women, 3 women, and 4 women and get the total ways that a committee can have at least 2 women.

 

19950 + 17100 + 4845 = 41895 ways that there will be at least 2 women in the committee


5 0
3 years ago
Multiplying Powers with the Same Base
Luda [366]

Applying the product rule of exponents, each product of powers are matched with its simplified expression as:

1. 5 \times 5^3 = 5^4

2. 5 \times 5^3 = 5^4

3. 5^{-3} \times 5^{-3} = \frac{1}{5^6}

4. 5^{-4} \times 5^{4} \times 5^0 = 5^0

5. 5^{7} \times 5^{3} = 5^{10}

To multiply the powers having the same base, we will apply the product rule for exponents.

<h3>What is the Product Rule for Exponents?</h3>
  • Base on the product rule for exponents, we have, a^m \times a^n = a^{m + n} = a^{mn}.
  • In order to find the products of two given numbers that have the same base, the exponents would be added together.

1. 5^6 \times 5^{-4

Add the exponents together

5^6 \times 5^{-4} = 5^{(6) + (-4)}

5^6 \times 5^{-4} = 5^2

2. 5 \times 5^3

Add the exponents together

5 \times 5^3 = 5^{(1 + 3)}

5 \times 5^3 = 5^4

3. 5^{-3} \times 5^{-3}

Add the exponents together

5^{-3} \times 5^{-3} = 5^{(-3) + (-3)

5^{-3} \times 5^{-3} = 5^{-6

5^{-3} \times 5^{-3} = \frac{1}{5^6}

4. 5^{-4} \times 5^{4} \times 5^0

Add the exponents together

5^{-4} \times 5^{4} \times 5^0 = 5^{(-4) + (4) + (0)

5^{-4} \times 5^{4} \times 5^0 = 5^0

5. 5^{7} \times 5^{3}

Add the exponents together

5^{7} \times 5^{3} = 5^{10}

In summary, applying the product rule of exponents, each product of powers are matched with its simplified expression as:

1. 5 \times 5^3 = 5^4

2. 5 \times 5^3 = 5^4

3. 5^{-3} \times 5^{-3} = \frac{1}{5^6}

4. 5^{-4} \times 5^{4} \times 5^0 = 5^0

5. 5^{7} \times 5^{3} = 5^{10}

Learn more about product rule of exponents on:

brainly.com/question/847241

5 0
2 years ago
15+8x=5y answer please
alexira [117]
15+8x+5y
8x+5y-15
x+5/8y-15/8

There is no solution
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3 years ago
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I NEEEDDDDD HELPPPP PLEASEEEEE!!!!
SCORPION-xisa [38]

Answer:

30.651

Step-by-step explanation:

hope it works!! :D

7 0
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