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Elden [556K]
3 years ago
15

In the equation, what is the value of x?

Mathematics
1 answer:
djyliett [7]3 years ago
5 0

Answer:

A) -7/9

Step-by-step explanation:

                     -4(x + 4x) = 2(2 - x) + 10

                     -4x - 16x = 4 - 2x + 10

-4x -  16x + 2x - 4 - 14 = 0

                         -18x -14 = 0

                               -18x = 14

                                   x = - 14/18

                                    x = - 7/9

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Because of their connection with secant​ lines, tangents, and instantaneous​ rates, limits of the form ModifyingBelow lim With h
Gre4nikov [31]

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\dfrac{1}{2\sqrt{x}}

Step-by-step explanation:

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

f(x+h) = \sqrt{x+h} = (x+h)^{\frac{1}{2}}

We use binomial expansion for (x+h)^{\frac{1}{2}}

This can be rewritten as

[x(1+\dfrac{h}{x})]^{\frac{1}{2}}

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From the expansion

(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}+\ldots

Setting x=\dfrac{h}{x} and n=\frac{1}{2},

(1+\dfrac{h}{x})^{\frac{1}{2}}=1+(\dfrac{h}{x})(\dfrac{1}{2})+\dfrac{\frac{1}{2}(1-\frac{1}{2})}{2!}(\dfrac{h}{x})^2+\tldots

=1+\dfrac{h}{2x}-\dfrac{h^2}{8x^2}+\ldots

Multiplying by x^{\frac{1}{2}},

x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}=x^{\frac{1}{2}}+\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots

x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}=\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots

\dfrac{x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}}{h}=\dfrac{1}{2x^{\frac{1}{2}}}-\dfrac{h}{8x^{\frac{3}{2}}}+\ldots

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