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3 years ago

Simplify (−2a^2b^4)^3

Mathematics

1 answer:

padilas [110]3 years ago

3 0

Answer:

- $-\frac{8a^6}{b^12}$

Step-by-step explanation:

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A farmer sells 8.7 kilograms of pears and apples at the farmer's market. 2 5 of this weight is pears, and the rest is apples. Ho

Ivenika [448]

Answer: 5.22kg

Step-by-step explanation:

From the question, a farmer sells 8.7 kilograms of pears and apples at the market. Of the pears and apples sold, 2/5 of this weight is pears, and the rest is apples. To calculate the kilograms of apple sold goes thus:

Total kilograms sold = 8.7kg

Fraction of pears weight = 2/5

Pears kilograms sold = 2/5 × 8.7

= 0.4 × 8.7

= 3.48kg

Since the kilograms of pears sold is , 3.48kg, to get the kilograms of apple sold, we subtract 3.48kg from 8.7kg. This will be:

= 8.7kg - 3.48kg

= 5.22kg

8 0

3 years ago

Evaluate the expression when b=4/5 and c=-3/8<br><br> 2c-b

timurjin [86]

Answer:

-1/20

Step-by-step explanation:

2(3/8) - 4/5

6/8 - 4/5

30/40 - 32/40

-2/40

-1/20

6 0

3 years ago

Someone please find the constant proportionality!!!

Andru [333]

Answer:

Possible - 1 and 0

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8 0

3 years ago

What is the slope of the following equation?

777dan777 [17]

Answer: A. -3

Step-by-step explanation: The slope is the coefficent of x, this can be read as -3/1, down 3 over 1.

6 0

3 years ago

A collection of 20 coins made up of only nickels, dimes, and quarters has a total value of $3.35. If the dimes were nickels, the

Kay [80]

Answer: The required number of quarters in the collection is 11.

Step-by-step explanation: Given that a collection of 20 coins made up of only nickels, dimes and quarters has a total value of $3.35.

If the dimes were nickels, the nickels were quarters and the quarters were dimes, the collection of coins would have a total value of $2.75.

We are to find the number of quarters in the collection.

Let x, y and z represents the number of nickels, dimes and quarters respectively in the collection.

We will be using the following values of nickels, dimes and quarters in form of dollar :

1 nickel = $ 0.05, 1 dime = $ 0.10 and 1 quarter = $0.25.

Then, according to the given information, we have

$x+y+z=20\\\\\Rightarrow x=20-y-z~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\0.05x+0.10y+0.25z=3.35\\\\\Rightarrow 5x+10y+25z=335\\\\\Rightarrow x+2y+5z=67~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)\\\\0.05y+0.25x+0.10z=2.75\\\\\Rightarrow 5y+25x+10z=275\\\\\Rightarrow y+2z+5x=55~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)$

Substituting the value of x from equation (i) in equations (ii) and (iii), we have

$(20-y-z)+2y+5z=67\\\\\Rightarrow y+4z=47\\\\\Rightarrow y=47-4z~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iv)$

and

$y+2z+5(20-y-z)=55\\\\\Rightarrow -4y-3z=-45\\\\\Rightarrow y=\dfrac{45-3z}{4}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(v)$

Comparing the values of y from equations (iv) and (v), we get

$47-4z=\dfrac{45-3z}{4}\\\\\Rightarrow 188-16z=45-3z\\\\\Rightarrow 16z-3z=188-45\\\\\Rightarrow 13z=143\\\\\Rightarrow z=\dfrac{143}{13}\\\\\Rightarrow z=11.$

Thus, the required number of quarters in the collection is 11.

7 0

3 years ago