3 years ago

Find the area and perimeter of the triangle

Mathematics

1 answer:

LekaFEV [45]3 years ago

3 0

Answer:

Step-by-step explanation:

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Neptune is an average distance of 4.5 × 10^9 km from the Sun. Estimate the length of the Neptunian year using the fact that the

zlopas [31]

Answer:

164.32 earth year

Step-by-step explanation:

distance of Neptune, Rn = 4.5 x 10^9 km

distance of earth, Re = 1.5 x 10^8 km

time period of earth, Te = 1 year

let the time period of Neptune is Tn.

According to the Kepler's third law

T² ∝ R³

$\left ( \frac{T_{n}}{T_{e}} \right )^{2}=\left ( \frac{R_{n}}{R_{e}} \right )^{3}$

$\left ( \frac{T_{n}}{1} \right )^{2}=\left ( \frac{4.5\times10^{9}}}{1.5\times10^{8}}} \right )^{3}$

Tn = 164.32 earth years

Thus, the neptune year is equal to 164.32 earth year.

Step-by-step explanation:

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3 years ago

What is the average rate of change of the function over the interval x = 0 to x = 5?

AURORKA [14]

$\bf slope = m = \cfrac{rise}{run} \implies \cfrac{ f(x_2) - f(x_1)}{ x_2 - x_1}\impliedby \begin{array}{llll} average~rate\\ of~change \end{array}\\\\[-0.35em] \rule{34em}{0.25pt}\\\\ f(x)= 2x^2-1\qquad \begin{cases} x_1=0\\ x_2=5 \end{cases}\implies \cfrac{f(5)-f(0)}{5-0} \\\\\\ \cfrac{[2(5)^2-1]~~-~~[2(0)^2-1]}{5}\implies \cfrac{50-(-1)}{5}\implies \cfrac{50+1}{5}\implies \cfrac{51}{5}\implies 10\frac{1}{5}$

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3 years ago

Find the area and perimeter of the triangle ​

Find the area and perimeter of the triangle