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eimsori [14]
3 years ago
5

the area of a rectangle is 70 square inches the length of the rectangle is 3 inches longer than the width wich equation models t

his situation
Mathematics
2 answers:
dezoksy [38]3 years ago
8 0

Answer:

x² - 3x - 70 = 0 models this situation.

Step-by-step explanation:

Area of a rectangle = 70 inches²

Let the length of the rectangle = x inch

Since Length of the rectangle is 3 inches longer than the width of the rectangle.

Then width of the rectangle = (x - 3) inches

Now area = Length × width

70 = x(x - 3)

x² - 3x = 70

x² - 3x - 70 = 0

Therefore, x² - 3x - 70 = 0 models this situation.

Greeley [361]3 years ago
6 0

Answer:3x+x^2

Step-by-step explanation:

Area= length*width

=(3+x) *x

=3x+x^2

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What is the greatest<br> common factor (GCF) of 16<br> and 21
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1

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Let's start by prime factorizing (finding the prime factors) of 16 and 21:

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60=5k is the answer for your question
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3 years ago
Help me simplify 7x-(x-y) and 9a-(2b-a)
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Solve the equation <img src="https://tex.z-dn.net/?f=%2835%20x%5E%7B4%7D%20y%2B14%20x%5E%7B5%7D%20y-2%20y%5E%7B3%7D-4x%20y%5E%7B
jeka94
\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0

M_y=35x^4+14x^5-6y^2-12xy^2
N_x=35x^4-6y^2

\dfrac{N_x-M_y}N=\dfrac{-14x^5+12xy^2}{7x^5-6xy^2}=-2

This suggests an integrating factor depending on x only is possible, and given by

\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^{2x}

Distributing across the ODE, we end up with

\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^{2x}}_{M^*}\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^{2x}}_{N^*}\,\mathrm dy=0

The equation is now exact, with

{M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^{2x}

Now we find the solution:

F_x=M^*
F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^{2x}\,\mathrm dx
F=(7x^5y-2xy^3)e^{2x}+f(y)

F_y=N^*
(7x^5-6xy^2)e^{2x}+f'(y)=(7x^5-6xy^2)e^{2x}
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The general solution is then

F(x,y)=(7x^5y-2xy^3)e^{2x}=C
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3 years ago
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