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nika2105 [10]
3 years ago
14

The price of milk increased from $2 to $2.50. What is the percent of change for the price of milk?

Mathematics
1 answer:
Alina [70]3 years ago
5 0

25.00%

Or

25c

Have a Merry Christmas!

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SOMEONE ANSWER THIS QUESTION BEFORE I GO CRAZY!!!
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Pls help w this it’s due at 9
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I worked it out on paper so i didn’t have to type hope this helps

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There are two fields whose total area is 56 square yards. One field produces grain at the rateof34bushel per square yard; the ot
fenix001 [56]

Answer:

the first field (rate 3/4) has 32 square yards and the second field (rate 2/3) has 24 square yards.

Step-by-step explanation:

With the statement we can make a system of 2x2 equations, where:

"x" is the area of the first field

"y" is the area of the second field

However,

x + y = 56 => x = 56 - y

3/4 * x + 2/3 * y = 40

replacing we have:

3/4 * (56 - y) + 2/3 * y = 40

42 - 3/4 * y + 2/3 * y = 40

-0.0833 * y = 40 - 42

y = -2 / -0.0833

y = 24

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Can someone help me with this?
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There are 5 equal sections on a spinner. Buzz wants to try to get the Letter D, which is 1/5 of the equal sectioned spinners.

In a 6 sided cube, 2, 4, & 6, are the even numbers, or 3/6 or 1/2 of the numbers are even.

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When temperature is zero degree Celsius, the Fahrenheit temperature is 32. When the Celsius temperature is 100, the correspondin
7nadin3 [17]

Answer:

\\ y = 1.8(x) + 32 or \\ y = \frac{9}{5}(x) + 32

or equivalently:

\\ F = 1.8(C) + 32 or \\ F = \frac{9}{5}(C) + 32

Step-by-step explanation:

To express the Fahrenheit temperature <em>as a linear function of the Celsius temperature</em>, F(c), we can proceed as follows.

We can use here <em>the two-point form</em> <em>equation</em> of a line:

\\ y-y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x-x_1) [1]

We are asked to express the <em>Fahrenheit temperature</em> as a function of <em>Celsius temperature</em>, so the independent variable, in this case, is <em>x</em> (Celsius temperature) and the dependent variable is <em>y</em> (Fahrenheit temperature).

When temperature is zero degree Celsius (\\x_1 = 0), the Fahrenheit temperature is 32 (\\y_1 = 32).

When the Celsius temperature is 100 (\\x_2 = 100), the corresponding Fahrenheit temperature is 212 (\\y_2 = 212).

Then, using [1], we have:

\\ y-32 = \frac{212 - 32}{100 - 0}(x-0)

\\ y-32 = \frac{180}{100}(x)

\\ y-32 = 1.8(x).

It could be also be written as:

\\ y-32 = \frac{18}{10}(x) = \\ y-32 = \frac{9}{5}(x), as it commonly appears in books.

Then <em>the Fahrenheit temperature express as a linear function of the Celsius temperature, F(c</em>) is ( solving the equation for <em>y </em>) :

\\ y = 1.8(x) + 32 or \\ y = \frac{9}{5}(x) + 32.

Or equivalently:

\\ F = 1.8(C) + 32 or \\ F = \frac{9}{5}(C) + 32

We can check this using the given values from the question:

For 0 Celsius degrees, the Fahrenheit temperature is:

\\ y = 1.8(0) + 32 = 32 Fahrenheit degrees.

For 100 Celsius degrees, the Fahrenheit temperature is:

\\ y = 1.8(100) + 32 = 180 + 32 = 212 Fahrenheit degrees.

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