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4 years ago

Solve the following equation for W. A= 2(L + W)

Mathematics

2 answers:

Ray Of Light [21]4 years ago

7 0

Answer:

A/2 - L = W

Step-by-step explanation:

Step 1: Write equation

A = 2(L + W)

Step 2: Solve for <em>W</em>

Distribute 2: A = 2L + 2W
Subtract 2L on both sides: A - 2L = 2W
Divide both sides by 2: A/2 - L = W

expeople1 [14]4 years ago

5 0

A/2-L=W just remember PEMDAS and you should be fine, hope this helps

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Gre4nikov [31]

Answer:

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

Step-by-step explanation:

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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 $n=\frac{1}{2}$ ,

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

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

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

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

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

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

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3 years ago

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Solnce55 [7]

Answer:

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Step-by-step explanation:

Step 1: Write out expression

6x² - 9x + 3 - (8x² + 7x - 1)

Step 2: Distribute negative

6x² - 9x + 3 - 8x² - 7x + 1

Step 3: Combine like terms (x²)

-2x² - 9x + 3 - 7x + 1

Step 4: Combine like terms (x)

-2x² - 16x + 3 + 1

Step 5: Combine like terms (constants)

-2x² - 16x + 4

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