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nignag [31]
3 years ago
10

Which ordered pair (p,r) is the solution to the given system of linear equations?

Mathematics
1 answer:
Misha Larkins [42]3 years ago
8 0

Answer:

(p,r) = (1/3, 2/9)

Step-by-step explanation:

Here, we want to solve a system of equations

We can rewrite the second equation by dividing through by 2

So we have;

4p + 3r = 2

and

5p - 3r = 1

Add both equations:

9p = 3

p = 3/9

p = 1/3

Recall ;

5p - 3r = 1

3r = 5p - 1

Substitute the value or p here

3r = 5(1/3)-1

3r = 5/3 - 1

3r = 2/3

r = 2/9

So we have the solution set as;

(p,r) = (1/3 , 2/9)

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F(x) = 3x + x3
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Answer:

Please check the explanation.

Step-by-step explanation:

Given

f(x) = 3x + x³

Taking differentiate

\frac{d}{dx}\left(3x+x^3\right)

\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'

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solving

\frac{d}{dx}\left(3x\right)

\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'

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\mathrm{Apply\:the\:common\:derivative}:\quad \frac{d}{dx}\left(x\right)=1

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now solving

\frac{d}{dx}\left(x^3\right)

\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}

=3x^{3-1}

=3x^2

Thus, the expression becomes

\frac{d}{dx}\left(3x+x^3\right)=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^3\right)

                    =3+3x^2

Thus,

f'(x) = 3 + 3x²

Given that f'(x) = 15

substituting the value  f'(x) = 15 in f'(x) = 3 + 3x²

f'(x) = 3 + 3x²

15 =  3 + 3x²

switch sides

3 + 3x² = 15

3x² = 15-3

3x² = 12

Divide both sides by 3

x² = 4

\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=\sqrt{4},\:x=-\sqrt{4}

x=2,\:x=-2

Thus, the value of x​ will be:

x=2,\:x=-2

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