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tamaranim1 [39]
3 years ago
5

Give 0.25ml twice a day for 8 days what is the quantity to dispense

Mathematics
1 answer:
bazaltina [42]3 years ago
8 0

Answer:

4 mL

Step-by-step explanation:

Multiply .25 by 2 because you are dispensing it twice a day.

.25 x 2 = .5, so you dispense .5 mL a day.

Now, multiply .5 mL a day x 8 days.

.5 x 8 = 4

Therefore, you should dispense 4 mL.

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Vinvika [58]

Answer:

b

Step-by-step explanation:

6 0
3 years ago
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The rock hard company is going to transport truckloads of stones to a highway project. It will cost $4200 to rent trucks, and it
Darya [45]

Answer:

c =4200+275s

Step-by-step explanation:

C is the total cost

4200 is how much to rent the trucks

275 is the price for each ton

s is how many tons of stone in being transported

4 0
3 years ago
What is the simplest form of this expression? (x + 7)(3x − 8)
Komok [63]
Answer: 3x^+13x-56
Steps
1). Multiply the parentheses
(X+7)(3x-8)
=X x 3x -8x+7 x 3x -7 x 8
2). Multiply
3x^-8x + 21x - 56
3). Do like terms
3x^+13x - 56
4). Solution
[3x^+13x - 56]

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4 0
3 years ago
What is the value of y if 13/9 = y/18
Maslowich

Answer:

y=26

Step-by-step explanation:

1. Swap sides

\frac{13}{9}=\frac{y}{18}

Swap sides:

\frac{y}{18}=\frac{13}{9}

2. Isolate the y

\frac{y}{18}=\frac{13}{9}

Multiply to both sides by 18:

\frac{y}{18}\cdot 18=\frac{13}{9}\cdot 18

Group like terms:

\frac{1}{18}\cdot 18y=\frac{13}{9}\cdot 18

Simplify the fraction:

y=\frac{13}{9}\cdot 18

Multiply the fractions:

y=\frac{13\cdot 18}{9}

Simplify the arithmetic:

y=26

---------------------------

Why learn this:

  • Linear equations cannot tell you the future, but they can give you a good idea of what to expect so you can plan ahead. How long will it take you to fill your swimming pool? How much money will you earn during summer break? What are the quantities you need for your favorite recipe to make enough for all your friends?
  • Linear equations explain some of the relationships between what we know and what we want to know and can help us solve a wide range of problems we might encounter in our everyday lives.

---------------------

Terms and topics

  • Linear equations with one unknown

The main application of linear equations is solving problems in which an unknown variable, usually (but not always) x, is dependent on a known constant.

We solve linear equations by isolating the unknown variable on one side of the equation and simplifying the rest of the equation. When simplifying, anything that is done to one side of the equation must also be done to the other.

An equation of:

ax+b=0

in which a and b are the constants and  x is the unknown variable, is a typical linear equation with one unknown. To solve for x in this example, we would first isolate it by subtracting b from both sides of the equation. We would then divide both sides of the equation by a resulting in an answer of:

x = -b/a

8 0
2 years ago
What is the true solution to 3 In 2+In 8 = 2 In(4x)?<br> 60f<br> Š O OOO<br> TNT 00
alina1380 [7]

Answer:

  x = 2

Step-by-step explanation:

Taking antilogs, you have ...

  2³ × 8 = (4x)²

  64 = 16x²

  x = √(64/16) = √4

  x = 2 . . . . . . . . (the negative square root is not a solution)

___

You can also work more directly with the logs, if you like.

  3·ln(2) +ln(2³) = 2ln(2²x) . . . . . . . . . . . write 4 and 8 as powers of 2

  3·ln(2) +3·ln(2) = 2(2·ln(2) +ln(x)) . . . . use rules of logs to move exponents

  6·ln(2) = 4·ln(2) +2·ln(x) . . . . . . . . . . . . simplify

  2·ln(2) = 2·ln(x) . . . . . . . . . . . subtract 4ln(2)

  ln(2) = ln(x) . . . . . . . . . . . . . . divide by 2

  2 = x . . . . . . . . . . . . . . . . . . . take the antilogs

4 0
2 years ago
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