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Bezzdna [24]
3 years ago
10

Solve the following equation for x.

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

Solve for x (isolate/get x by itself)

\frac{1}{4}x-7=\frac{3}{8}x-5    Add 5 on both sides

\frac{1}{4}x-7+5=\frac{3}{8}x-5+5

\frac{1}{4}x-2=\frac{3}{8}x   Subtract 1/4x on both sides

\frac{1}{4}x-\frac{1}{4}x-2=\frac{3}{8}x-\frac{1}{4}x

-2=\frac{3}{8}x-\frac{1}{4}x   Make the denominators the same of the fractions in order to combine them. Multiply 2 to the top and bottom of 1/4x

-2=\frac{3}{8}x-\frac{2}{8}x

-2=\frac{1}{8}x    Multiply 8 on both sides

-2(8)=(8)\frac{1}{8}x

-16 = x

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Answer:

\dfrac{47}{60} sq. units.

Step-by-step explanation:

The given function is

h(x)=\dfrac{1}{7-x}

We need to find the area between x-axis and the given function from x=2 to x=5.

Left Riemann sum formula of area:

Area=\sum_{n=0}^{N-1}f(x_n)(\Delta x_n)

For given question,

Area=\sum_{n=2}^{5-1}f(x_n)(\Delta x_n)

Area=\sum_{n=2}^{4}f(x_n)(\Delta x_n)

Area=f(x_2)(3-2)+f(x_3)(4-3)+f(x_4)(5-4)

Now,

Area=\dfrac{1}{7-2}\times (1)+\dfrac{1}{7-3}\times (1)+\dfrac{1}{7-4}\times (1)

Area=\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{3}

Area=\dfrac{12+15+20}{60}

Area=\dfrac{47}{60}

Therefore, the required area is \dfrac{47}{60} sq. units.

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