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3 years ago

If you had 4 quarters and 8 nickels how much money would you have?

Mathematics

2 answers:

spin [16.1K]3 years ago

6 0

Answer and Step-by-step explanation:

A Quarter is equal to 25 cents.

A Nickel is equal to 5 cents.

We have 4 quarters and 8 nickels, so times this amount of coins to their values.

4 × 25 = 100

8 × 5 = 40

Now, we add these together.

140.

This is 140 cents.

After the 100 mark, it turns into a dollar (because there is 100 pennies (cents) in 1 dollar).

So that means the total amount of money we have is One dollar and Forty cents ($1.40)

#teamtrees #PAW (Plant And Water)

fenix001 [56]3 years ago

6 0

Answer:

$1.40

Step-by-step explanation:

4 quarters

1 quarter = 25 cents

Therefore, 4 quarters = 1/25 x 4/x

multiply 4 by 25, since 1 quarter = 25 cents

4 x 25 = 100

4 quarters = 100 cents

which equals $1

8 nickels

1 nickle = 5 cents

Therefore, 8 nickles = 1/5 x 8/x

multipy 8 by 5, since 1 nickel = 5cents

8 x 5 = 40

8 nickels = 40 cents

So add both 40 cents and 100 cents, which equals 140 cents.

But you still have to change the cents to dollars.

Which is 100 cents = $1

Add 40 cents

= $1.40

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3 years ago

What is (-33x-78)+(5x8x9x10)divided by9.75

kherson [118]

Answer:

6,174

Step-by-step explanation:

(-33 * - 78) + (5 * 8 * 9 * 10)

Remove the brackets:

-33 * - 78 + 5 * 8 * 9 * 10

Solve like so:

-33 * - 78 = 2,574

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3 years ago

An elephant at the zoo maintains her weight by eating 2.5% of her body weight in food each day. If she maintains a constant weig

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Answer:

2100

Step-by-step explanation:

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3 years ago

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mylen [45]

Answer:

56.39 nm

Step-by-step explanation:

In order to have constructive interference total optical path difference should be an integral number of wavelengths (crest and crest should be interfered). Therefore the constructive interference condition for soap film can be written as,

$2t=(m+\frac{1}{2} ).\frac{\lambda}{n}$

where λ is the wavelength of light and n is the refractive index of soap film, t is the thickness of the film, and m=0,1,2 ...

Please note that here we include an additional 1/2λ phase shift due to reflection from air-soap interface, because refractive index of latter is higher.

In order to have its longest constructive reflection at the red end (700 nm)

$t_1=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_1=\frac{1}{2} .\frac{700}{(2)*(1.33)}\\ \\ t_1=131.58\ nm$

Here we take m=0.

Similarly for the constructive reflection at the blue end (400 nm)

$t_2=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_2=\frac{1}{2} .\frac{400}{(2)*(1.33)}\\ \\ t_2=75.19\ nm$

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$t_1-t_2=131.58-75.19=56.39 \ nm$

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2 years ago