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3 years ago

What the value of x plz

Mathematics

1 answer:

Gwar [14]3 years ago

3 0

Well, the answer to this question is 55 degrees.
In this whole line intercept, the total number of degrees is 360 degrees and if 55 plus 55 equals to 110, then to solve the missing two angles, we can just subtract 110 from 360 which equals to 250 and lastly, we have to divide 250 by 2 which is 125.

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Ginny has two vegetable gardens.In the second garden,Ginny grows tomatoes every 3 years.This year,Ginny grows tomatoes in both g

kherson [118]

So it would be in six years because if you do it, it would be
garden 1: 2,4,6,8
garden 2: 3,6,9
and the lowest common multiple is 6 so in 6 years is when it will happen again

7 0

3 years ago

y=c1e^x+c2e^−x is a two-parameter family of solutions of the second order differential equation y′′−y=0. Find a solution of the

vagabundo [1.1K]

The general form of a solution of the differential equation is already provided for us:

$y(x) = c_1 \textrm{e}^x + c_2\textrm{e}^{-x},$

where $c_1, c_2 \in \mathbb{R}$ . We now want to find a solution $y$ such that $y(-1)=3$ and $y'(-1)=-3$ . Therefore, all we need to do is find the constants $c_1$ and $c_2$ that satisfy the initial conditions. For the first condition, we have: $y(-1)=3 \iff c_1 \textrm{e}^{-1} + c_2 \textrm{e}^{-(-1)} = 3 \iff c_1\textrm{e}^{-1} + c_2\textrm{e} = 3.$

For the second condition, we need to find the derivative $y'$ first. In this case, we have:

$y'(x) = \left(c_1\textrm{e}^x + c_2\textrm{e}^{-x}\right)' = c_1\textrm{e}^x - c_2\textrm{e}^{-x}.$

Therefore:

$y'(-1) = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e}^{-(-1)} = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e} = -3.$

This means that we must solve the following system of equations:

$\begin{cases}c_1\textrm{e}^{-1} + c_2\textrm{e} = 3 \\ c_1\textrm{e}^{-1} - c_2\textrm{e} = -3\end{cases}.$

If we add the equations above, we get:

$\left(c_1\textrm{e}^{-1} + c_2\textrm{e}\right) + \left(c_1\textrm{e}^{-1} - c_2\textrm{e} \right) = 3-3 \iff 2c_1\textrm{e}^{-1} = 0 \iff c_1 = 0.$

If we now substitute $c_1 = 0$ into either of the equations in the system, we get:

$c_2 \textrm{e} = 3 \iff c_2 = \dfrac{3}{\textrm{e}} = 3\textrm{e}^{-1.}$

This means that the solution obeying the initial conditions is:

$\boxed{y(x) = 3\textrm{e}^{-1} \times \textrm{e}^{-x} = 3\textrm{e}^{-x-1}}.$

Indeed, we can see that:

$y(-1) = 3\textrm{e}^{-(-1) -1} = 3\textrm{e}^{1-1} = 3\textrm{e}^0 = 3$

$y'(x) =-3\textrm{e}^{-x-1} \implies y'(-1) = -3\textrm{e}^{-(-1)-1} = -3\textrm{e}^{1-1} = -3\textrm{e}^0 = -3,$

which do correspond to the desired initial conditions.

3 0

3 years ago

Solve for x,,,3x-1/4=-5/11

Neko [114]

$\dfrac{3x-1}{4}=-\dfrac{5}{11}\ \ \ \ |cross\ multiply\\\\11(3x-1)=4\cdot(-5)\\\\33x-11=-20\ \ \ \ |add\ 11\ to\ both\ sides\\\\33x=-9\ \ \ \ |divide\ both\ sides\ by\ 33\\\\x=-\dfrac{9}{33}\\\\\boxed{x=-\dfrac{3}{11}}$

4 0

4 years ago

Simplify 4x2y + 12z2 + 6x2y – 2z.

yaroslaw [1]

Divide all the terms by 2:
2xy + 12z + 3xy - z

5 0

3 years ago

Read 2 more answers

Pleaseee help me quicly

leonid [27]

Answer:

Q.4=2x^2

Q.5=6l^2

may be this is helpful.

8 0

2 years ago