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3 years ago

15

Find the missing side b in a right triangle where a=8 and c=18

1 answer:

DochEvi [55]3 years ago

4 0

Using Pythagorean theorem c^2=a^2 + b^2, we can find the missing side b.
Put values in equation.
18^2 = 8^2 + b^2
324=64+b^2
Subtract 64 from both sides
260= b^2
Take square root.
2√65=b
or
b=16.12

Answer: b=2√65 Radical form. or b=16.12 Decimal form.

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Please help! Solve the equation. Check your solution, if possible. #4 (10 points) picture is included) please help!

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N=7.5 it's a real solution

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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]

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

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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

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1. If a height of each of the bars in a bar graph get taller as you go from the far left to the far right, the data are:

Alina [70]

Answer:

1. Skewed to left (negative)

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Step-by-step explanation:

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goblinko [34]

The answer should be B

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