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3 years ago

What is the solution to the system of equations below?

Mathematics

2 answers:

Monica [59]3 years ago

6 0

Answer:

Step-by-step explanation:

$\left\{\begin{array}{ccc}y=\dfrac{1}{2}x-4&(1)\\y=-2x-9&(2)\end{array}\right\\\\\text{substitute (1) to (2):}\\\\\dfrac{1}{2}x-4=-2x-9\qquad\text{multiply both sides by 2}\\\\x-8=-4x-18\qquad\text{add 8 to both sides}\\\\x=-4x-10\qquad\text{add 4x to both sides}\\\\5x=-10\qquad\text{divide both sides by 5}\\\\x=-2\\\\\text{put it to (2):}\\\\y=-2(-2)-9\\\\y=4-9\\\\y=-5$

Romashka-Z-Leto [24]3 years ago

4 0

4(y= 1/2x -4)

4y = 2x -16
y= -2x -9

5y = -25

Y= -5

So since there is only one answer with y= -5 the answer is A

Hope this helps!

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Answer:

Step-by-step explanation:

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3 years ago

Use linear approximation, i.e. the tangent line, to approximate √16.4 as follows: Let f(x)=√x. Find the equation of the tangent

Ivahew [28]

Answer:

$L(x)=\frac{1}{8}x+2\\ L(16.4)=4.05$

Step-by-step explanation:

The equation for a tangent line of f(x) in the point (a,f(a)) can be calculated as:

L(x) = f(a) + f'(a)(x-a)

Where L(x) is also call a linear approximation and f'(a) is the value of the derivative of f(x) in (a,f(a)).

So, the derivative of f(x) is:

$f(x)=\sqrt{x} \\f'(x)=\frac{1}{2\sqrt{x} }$

Then, to find the linear approximation we are going to use the point (16, f(16)). So a is 16 and f(a) and f'(a) are calculated as:

$f(16)=\sqrt{16}=4\\f'(16)=\frac{1}{2\sqrt{16} }=\frac{1}{8}$

Then, replacing the values, we get that the equation of the tangent line in (16,4) is:

$L(x)=4+\frac{1}{8}(x-16)\\L(x) = \frac{1}{8}x+2$

Finally, the approximation for $\sqrt{16.4}$ is:

$L(16.4)=\frac{1}{8}(16.4)+2\\ L(16.4)=4.05$

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Answer:

Step-by-step explanation:

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Step-by-step explanation:

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