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algol [13]
4 years ago
5

Marcus finds that (3x^2-2y^2+5x)+(4x^2+12y-7x)= 7x^2-10y^2-2x What error did Marcus make?

Mathematics
2 answers:
Yuri [45]4 years ago
8 0

we have

(3x^{2} -2y^{2}+5x)+(4 x^{2}+12y^{2}-7x)

Combine like terms

(3x^{2} -2y^{2}+5x)+(4 x^{2}+12y^{2}-7x)\\=(3x^{2}+4x^{2})+(-2y^{2}+12y^{2})+(5x-7x)\\=7x^{2} +10y^{2}-2x

therefore

<u>the answer is the option C</u>

He combined the terms -2y^{2} and 12y^{2} incorrectly

Juli2301 [7.4K]4 years ago
4 0
I believe the correct answer from the choices listed above is option C. The error that Marcus made would be that he <span>combined the terms –2y^2 and 12y^2 incorrectly. It should be 10y^2 when combined not -10y^2. Hope this answers the question.</span>
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Answer:

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The range of the function H(t), is [(10% + average), (average - 20%)]

Step-by-step explanation:

The domain of a function is the complete set of possible values of the independent variable.

For this question, the function is H(t), with the temperature, t, serving as the independent variables and H(t) the evidently dependent variable.

The domain of a function refers to all the possible independent variable values that will give corresponding real dependent variable values.

For this question, Alice's model has the probability for the occurrence of heart disease (in percents relative to the global average) at an area, H(t) varying with the temperature of that area in degree Celsius.

At a temperature of -5°C (the lowest temperature in the model), the probability is 10% above the average.

Then, the probability decreases with increase in temperature, taking a value 20% lower than the average when the temperature is at its highest of 30°C in the model.

So, temperature ranges from -5°C to 30°C and the probability for the occurrence of heart disease ranges from 10% above the average to 20% below the average.

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And the range of the function H(t), is [(10% + average), (average - 20%)]

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7 0
3 years ago
Read 2 more answers
What binomial do you have to add to the polynomial:
marshall27 [118]

Answer:

a)  -2x^2+5xy

b)  -3y^2+5xy

Step-by-step explanation:

<u>Part (a)</u>

Given polynomial  :2x^2+3y^2-5xy+1

The binomial that should be added to the given polynomial to get a polynomial that does not contain the variable x is:

-2x^2+5xy

(2x^2+3y^2-5xy+1)+(-2x^2+5xy)

=2x^2+3y^2-5xy+1-2x^2+5xy

=2x^2-2x^2+3y^2-5xy+5xy+1

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<u>Part (b)</u>

Given polynomial  :2x^2+3y^2-5xy+1

The binomial that should be added to the given polynomial to get a polynomial that does not contain the variable y is:

-3y^2+5xy

(2x^2+3y^2-5xy+1)+(-3y^2+5xy)

=2x^2+3y^2-5xy+1-3y^2+5xy

=2x^2+3y^2-3y^2-5xy+5xy+1

=2x^2+1

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