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3 years ago

How many midsegments does a triangle have?

Mathematics

2 answers:

katrin2010 [14]3 years ago

8 0

A triangle has three possible midsegments<span>, depending on which pair of sides is initially joined.</span>

GREYUIT [131]3 years ago

4 0

The midsegments the triangle have is three

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Write an equation that gives the overall equilibrium constant in terms of the equilibrium constants and . If you need to include

shutvik [7]

Answer:

The answer is " $k=k_1k_{2}^2$ "

Step-by-step explanation:

For point 1:

$4NHH_3 \ (g) +5O_2\ (g) \leftrightharpoons 4NO\ (g) +6H_2O \ (g)..................k_1\\\\$

For point 2:

$2 \times(2\ NO\ (o)+O_2 \ (g) \leftrightharpoons 2NO_2\ (g)) ..................K_2\\\\$

Adding $k_1$ equation and multiply by 2 in the equation $k_2$ so we get $\to 4NH_3\ (g) +7 O_2\ (g) \leftrightharpoons 4 NO_2\(g) +6H_2O\ (g)....... k\\\\k= k_1\times k_{2}^2\\\\k=k_1k_{2}^2\\\\$

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3 years ago

Please help asap! Will give brainliest! Please answer correctly!<br> No guessing!

kow [346]

Answer:

Let t be temperature

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Step-by-step explanation:

4 0

3 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

3 years ago

How do you find the volume of a pyramid?

amm1812

By knowing the volume of a triangle and that the triangle has all equal sides

4 0

3 years ago

Read 2 more answers

In a swimming class, 2/5 of the participants are girls. there are 24 boys in the class. find the total number of participants in

jeyben [28]

30 because 2/5 are girls so 3/5 are boys and 24/3 +8 so 1/5=8 so 8*5 + 40 kids

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8 0

3 years ago