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mart [117]
2 years ago
10

10 3/9- 10 2/5 Im to lazy to do it please help me out.

Mathematics
1 answer:
coldgirl [10]2 years ago
8 0

Answer:

\boxed{ \ -  \frac{3}{45} }

Step-by-step explanation:

10 3/9 - 10 2/5

we need to convert mixed fractions to improper fractions

10 \frac{3}{9}  =  \frac{10 \times 9 + 3}{9}  =  \frac{93}{9}

10 \frac{2}{5}  =  \frac{10 \times 5 + 2}{5}  =  \frac{52}{5}

Now, we have 93/9 - 52/5

We need to equate the denominators, before subtracting them.

\frac{93}{9}  -  \frac{52}{5}  =  \frac{93 \times 5}{45}  -  \frac{52 \times 9}{45}

\frac{465}{45}  -  \frac{468}{45}

\frac{ - 3}{45}

\boxed{ -  \frac{3}{45} }

The result of 10 3/9- 10 2/5 is - 3/45

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Write the integral that gives the length of the curve y = f (x) = ∫0 to 4.5x sin t dt on the interval ​[0,π​].
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Answer:

Arc length =\int_0^{\pi} \sqrt{1+[(4.5sin(4.5x))]^2}\ dx

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Step-by-step explanation:

The arc length of the curve is given by \int_a^b \sqrt{1+[f'(x)]^2}\ dx

Here, f(x)=\int_0^{4.5x}sin(t) \ dt interval [0, \pi]

Now, f'(x)=\frac{\mathrm{d} }{\mathrm{d} x} \int_0^{4.5x}sin(t) \ dt

f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( [-cos(t)]_0^{4.5x} \right )

f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left ( -cos(4.5x)+1 \right )

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Now, the arc length is \int_0^{\pi} \sqrt{1+[f'(x)]^2}\ dx

\int_0^{\pi} \sqrt{1+[(4.5sin(4.5x))]^2}\ dx

After solving, Arc length =9.75053

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