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Gekata [30.6K]
3 years ago
6

Subtract. (5x2−3x−2)−(−2x2−x+10) Express the answer in standard form. Enter your answer in the box.

Mathematics
2 answers:
Nimfa-mama [501]3 years ago
5 0

Answer:

7x^{2}-2x-12=0

Step-by-step explanation:

we know that

The standard form of the quadratic equation is equal to ax^{2} +bx+c=0

we have

(5x^{2}-3x-2)-(-2x^{2}-x+10)

eliminate the parenthesis

5x^{2}-3x-2+2x^{2}+x-10

group terms that contain the same variable

(5x^{2}+2x^{2})+(-3x+x)+(-2-10)

combine like terms

7x^{2}-2x-12=0

inna [77]3 years ago
4 0

Answer:

7x²-2x-12

Step-by-step explanation:

We have given an expression.

(5x²-3x-2)-(-2x²-x+10)

We have to subtract them.

ax²+bx+c is general form of an expression.

We have to transform above equation in general form of expression.

Open parentheses and distribute negative sign to second parentheses.

5x²-3x-2-(-2x²)-(-x)-(10)

5x²-3x-2+2x²+x-10

Collect the like terms

5x²+2x²-3x+x-2-10

Add like terms

7x²-2x-12 which is the answer.

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Is 124 a whole number
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A one-parameter family of solutions of the de p' = p(1 − p) is given below.
tensa zangetsu [6.8K]

Answer:

A solution curve pass through the point (0,4) when c_{1} = -\frac{4}{3}.

There is not a solution curve passing through the point(0,1).

Step-by-step explanation:

We have the following solution:

P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}

Does any solution curve pass through the point (0, 4)?

We have to see if P = 4 when t = 0.

P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}

4 = \frac{c_{1}}{1 + c_{1}}

4 + 4c_{1} = c_{1}

c_{1} = -\frac{4}{3}

A solution curve pass through the point (0,4) when c_{1} = -\frac{4}{3}.

Through the point (0, 1)?

Same thing as above

P(t) = \frac{c_{1}e^{t}}{1 + c_{1}e^{t}}

1 = \frac{c_{1}}{1 + c_{1}}

1 + c_{1} = c_{1}

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No solution.

So there is not a solution curve passing through the point(0,1).

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3 years ago
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