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natulia [17]
3 years ago
12

In the equation 3x/4=1/x the value of b is

Mathematics
1 answer:
erica [24]3 years ago
7 0

Answer:

Step-by-step explanation:

x=±\dfrac{2}{\sqrt{3}}

Step-by-step explanation:

We want to solve the equation

\dfrac{3x}{4}=\dfrac{1}{x}

We first have to assume that x\neq 0 since the value \dfrac{1}{x} is not defined for x\neq 0.

Now, since x divides the right hand side of the equation, we multiply the two sides of this equation by x to get that

x(\frac{3}{4}x)=x(\dfrac{1}{x})\quad\Rightarrow\quad \dfrac{3}{4}x^2=1

Hence, by dividing the equation by \dfrac{4}{3} it follows that

x^2=\dfrac{4}{3}

and so

x=\pm\dfrac{2}{\sqrt{3}}

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Nookie1986 [14]

Answer:

-10,000 / -2000, although 10,000/2000 gives the same answer.

Step-by-step explanation:

The equations -10,000/2000 and 10,000/-2000 are negative, and negative days do not make sense, thus eliminating those answers.

Also, she wants to descend 10,000 feet, so that makes it -10,000 int he equation.

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What's the converse of the statement "if it is snowing, then it is my birthday"
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3 years ago
In a school 50% of learner's are younger than 10 , 1/20 are 10years old and 1/10 are older than 10 but younger than 12, the rema
frosja888 [35]

Answer:

1/20=10 years old

In a school 50% of the students are younger than 10, 1/20 are 10 years old and 1/10 are older than 10 but younger than 12, the remaining 70 students are 12 years or older. How many students are 10 years old?

50%= younger than 10

1/20=10 years old

1/10=older than 10 younger than 12

70=12 years or older

50%- 10/20

1/20-1/20

1/10-2/20

10/20+1/20+2/20=13/20

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70 divided by 7=10

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2 years ago
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What differentiable functions have an arc length on the interval [a, b] given by the following integrals? Note that the answers
Dafna1 [17]

Answer:

a) \pm 8\frac{x^{5}}{5}+C

b) \pm 4 sin(2x) + C

Step-by-step explanation:

Given integrals are:

a) \int\limits^a_b {\sqrt{{1+64x^{8}} } \, dx ---(1)

b) \int\limits^a_b {\sqrt{1+64cos^{2}(2x)} } \, dx---- (2)

Standard form

                L= \int\limits^a_b {\sqrt{1+(f'(x))^{2}} } \, dx

<h3>Part A</h3>

compare (1) with standard form

[f'(x)]^{2} = 64x^{8}\\f'(x)=\pm 8x^{4}\\f(x)= \pm8\frac{x^{5}}{5}+C

<h3>Part B</h3>

Compare (2) with standard form

[f'(x)]^{2}=64cos^{2}(2x)\\f'(x)= \pm 8cos(2x)\\f(x)=\pm 8\frac{sin(2x)}{2}+C\\f(x)= \pm 4sin(2x)+C

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3 years ago
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alina1380 [7]
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