3 years ago

Find BC round to the nearest tenth

Mathematics

1 answer:

slega [8]3 years ago

3 0

Answer:

6.7

Step-by-step explanation:

Here, we can apply cosine law, $c^2=a^2+b^2-2abcos\theta$

We the plug in values: $(CB)^2+6^2+5^2-2*6*5*cos74$

and solve that CB is around 6.7

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Compute Δy and dy for the given values of x and dx = Δx. (Round your answers to three decimal places.) y = ex, x = 0, Δx = 0.4

kobusy [5.1K]

Answer:

0.492

Step-by-step explanation:

According to the given situation, the calculation of Δy and dy is shown below:-

$y = e^x, x = o, \Delta x = 0.6$

$dy = f'(x) = e^x$

$dy = e^x dx$

$= e^x (0.4)$

$dy = 0.4e^x$

$\Delta y = f(0.4) - f(0)$

$= e^{0.4} - e^0$

Now we use the scientific calculator or spreadsheet to determine the exponential value

= 1.491824698 - 1

= 0.491824698

= 0.492

Therefore for computing the Δy and dy we simply applied the above formula.

Hence, the answer is 0.4918

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