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Inga [223]
3 years ago
7

1. What is the solution of n^2-49=0?

Mathematics
1 answer:
Effectus [21]3 years ago
7 0

<u>Answer:</u>

<h2>1. a</h2><h2>2. b</h2><h2>3. b</h2><h2>4. b</h2><h2>5. 22</h2>

<u>Explanation:</u>

1.

n²-49=0

n=√49

n=7

2.

x²+64=0

x=√64

x=8

3.

if A=144x²

l = √A

l = √(144x²)

l = 12x

4.

b = (24×2)/3b

b = 48/3b

b = 16/b

b² = 16

b = √16

b = 4 in <em>(-4 is correct in algebra but there's no negative lengths in geometric shapes)</em>

5.

x²+z = 103

x² = 103-z

x = √(103-z)

for x = -9 and 9    (103-z) must equal 81

z = 103 - 81 = 22

x = √(103-22)

x = √81

x = -9 , 9

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Step-by-step explanation:

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3 years ago
There are some Nichols dimes and quarters in a large piggy bank for every two nickels there a sweetheart forever to Diane’s ther
grigory [225]

Answer:

There are 80 nickels, 120 dimes and 300 quarters in the piggy bank.

All coins in the piggy bank worth $91.

Step-by-step explanation:

Please consider the complete question:

There are some nickels, dimes, and quarters in a large piggy bank. For every 2 nickels there are 3 dimes. For every 2 dimes there are 5 quarters. There are 500 coins total. How many nickels, dimes, and quarters are in the piggy bank? Explain your reasoning. How much are the coins in the piggy bank worth?

Let n, d, q represent number of nickles, dimes and quarters respectively.

For every 2 nickels there are 3 dimes. We can represent this information in an equation as:

\frac{2}{n}=\frac{3}{d}...(1) or

n=\frac{2d}{3}...(1)

For every 2 dimes there are 5 quarters. We can represent this information in an equation as:

\frac{2}{d}=\frac{5}{q}...(2) or  

q=\frac{5d}{2}...(2)  

There are 500 coins total. We can represent this information in an equation as:

n+d+q=500...(3)

We will use substitution method to solve our given system.

Upon substituting equation (1) and equation (2) in equation (3), we will get:

\frac{2d}{3}+d+\frac{5d}{2}=500  

Multiply the equation by 6:

\frac{2d}{3}*6+d*6+\frac{5d}{2}*6=500*6  

2d*2+6d+5d*3=3000  

4d+6d+15d=3000  

25d=3000  

\frac{25d}{25}=\frac{3000}{25}  

d=120  

Therefore, there are 120 dimes in the piggy bank.

Upon substituting d=120 in equation (1), we will get:

n=\frac{2(120)}{3}

n=2(40)

n=80

Therefore, there are 80 nickels in the piggy bank.

Upon substituting d=120 in equation (2), we will get:

q=\frac{5d}{2}  

q=\frac{5(120)}{2}  

q=5(60)  

q=300  

Therefore, there are 300 quarters in the piggy bank.

Now, we need to find the value of all coins.

We know 1 quarter is worth $0.25, so 300 quarters will be worth \$0.25\times 300=\$75.

We know 1 dime is worth $0.10, so 120 dimes will be worth \$0.10\times 120=\$12.

We know 1 nickel is worth $0.05, so 80 nickels will be worth \$0.05\times 80=\$4.

\text{All coins worth}=\$75+\$12+\$4

\text{All coins worth}=\$91

Therefore, all coins in the piggy bank worth $91.

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