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2 years ago

F(x) = 3x - 2 f(x) = 0, x=?

Mathematics

2 answers:

podryga [215]2 years ago

8 0

Answer:

2/3 =x

Step-by-step explanation:

f(x) = 3x - 2

Let f(x) =0

0 = 3x-2

Add 2 to each side

2 = 3x-2+2

2 =3x

Divide by 3

2/3 =3x/3

2/3 =x

prohojiy [21]2 years ago

6 0

Answer:

f(x)=3x-2

0=3x-2

2=3x

2/3=x

x=2/3

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Marat540 [252]

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2 years ago

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Step-by-step explanation:

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3 years ago

A rectangle has a length this is three times its width. If the area of the rectangle is 27 square feet, what are the dimensions

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Step-by-step explanation:

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3 years ago

Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form ModifyingBelow lim With h

Gre4nikov [31]

Answer:

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

Step-by-step explanation:

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This can be rewritten as

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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 $x^{\frac{1}{2}}$ ,

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

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

The limit of this as $h\to 0$ is

$\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h}=\dfrac{1}{2x^{\frac{1}{2}}}=\dfrac{1}{2\sqrt{x}}$ (since all the other terms involve $h$ and vanish to 0.)

8 0

3 years ago

How do you solve this? and please walk me though it! plz!!

Sidana [21]

Every triangles 3 angles will always add up to 180 degrees

The reason this is true in your problem is because of the corresponding angles and supplementary angles.

Hope this helps

4 0

3 years ago