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Kay [80]
2 years ago
15

Please help solve this, I solved it on my own but I think my calculations are wrong. here is the problem,

Mathematics
2 answers:
denis-greek [22]2 years ago
5 0

Hi there! Hopefully this helps!

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

<h2>Answer: 12</h2><h2>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~</h2>

First we need to <em><u>substitute the value of the variable into the expression</u></em> and simplify.

b-(b+a×4c÷4) = 1 - (1 + -6 × 4(2) ÷ 4).

Now we need to solve 1 - (1 + -6 × 4(2) ÷ 4).

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

1 - (1 + -6 × 4(2) ÷ 4)

|

|  First, we cancel out 4 and 4.

\/

1 - (1 - 6 × 2)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Next, we multiply -6 and 2 to get -12.

1 - (1 - 12)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Then we subtract 12 from 1 to get -11.

1 - (-11)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The opposite of -11 is 11.

1 + 11.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

<h2 /><h2>Add 1 and 11 to get, you guessed it, 12!</h2>
Lostsunrise [7]2 years ago
5 0

Answer:

Answer: 12

Hope this helps

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Step-by-step explanation:

We are given the following in the question:

The completion of a chemical reaction is expressed by the equation

y = 300 - 4x -x^2

where y is  the number of seconds needed to complete the reaction and x is the temperature in degrees Celsius  at which the reaction occurs.

We have to find the temperature at which the reaction occurs, if the completion time of reaction is 160 seconds.

Putting y = 160

160 = 300 - 4x -x^2\\x^2 + 4x  - 140 = 0\\x^2 + 14x -10x -140 = 0\\x(x+14)-10(x+14)=0\\(x-10)(x+14) = 0\\(x-10) =0, (x+14) =0\\x = 10, x = -14

Thus, the reaction ca occur at 10 degree Celsius or -14 degree Celsius.

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<em><u>Answer:</u></em>

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<em><u>Step-by-step explanation:</u></em>

So the ratio of the NEW : ORIGINAL is just one side of the rectangle to the other side of the original one.

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