The sum of the angle measures of any triangle is

Mathematics

1 answer:

romanna [79]3 years ago

3 0

Let's say that y is the unknown
180=2x+3+5x+1+y
180=7x+4+y
180-7x-4=y
176-7x=y=third angle

You might be interested in

Point M is the midpoint of AB¯¯¯¯¯¯¯¯. Find the length of AB¯¯¯¯¯¯¯¯.

alina1380 [7]

Answer:

AB = 142

Step-by-step explanation:

If M is the midpoint, then AM = MB

6x+5 = 7x -6

Subtract 6x from each side

6x-6x+5 = 7x-6x -6

5 = x-6

Add 6 to each side

5+6 = x-6+6

11=x

We want to find AB

AB = 6x+5+7x-6

AB = 13x -1

= 13(11) -1

=143-1

= 142

6 0

3 years ago

Help with 30 please. thanks.

Svet_ta [14]

Answer:

See Below.

Step-by-step explanation:

We have the equation:

$\displaystyle y = \left(3e^{2x}-4x+1\right)^{{}^1\! / \! {}_2}$

And we want to show that:

$\displaystyle y \frac{d^2y }{dx^2} + \left(\frac{dy}{dx}\right) ^2 = 6e^{2x}$

Instead of differentiating directly, we can first square both sides:

$\displaystyle y^2 = 3e^{2x} -4x + 1$

We can find the first derivative through implicit differentiation:

$\displaystyle 2y \frac{dy}{dx} = 6e^{2x} -4$

Hence:

$\displaystyle \frac{dy}{dx} = \frac{3e^{2x} -2}{y}$

And we can find the second derivative by using the quotient rule:

$\displaystyle \begin{aligned}\frac{d^2y}{dx^2} & = \frac{(3e^{2x}-2)'(y)-(3e^{2x}-2)(y)'}{(y)^2}\\ \\ &= \frac{6ye^{2x}-\left(3e^{2x}-2\right)\left(\dfrac{dy}{dx}\right)}{y^2} \\ \\ &=\frac{6ye^{2x} -\left(3e^{2x} -2\right)\left(\dfrac{3e^{2x}-2}{y}\right)}{y^2}\\ \\ &=\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\end{aligned}$