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2 years ago

5

Triangel ABC has vertices A(-2,1) B(2,2), and C(2,y). The area of the triangle is 4 square units.

1 answer:

lozanna [386]2 years ago

8 0

Answer:

i wish i knew

Step-by-step explanation:

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What is the diameter of this circle if the circumference is 30 x inches?

ch4aika [34]

Answer:

<h2>10 inches</h2>

Step-by-step explanation:

Given the circumference= 30 in

we know that the formula for circumference is

$C= 2\pi r$

$30= 2*3.142*r\\\\r= \frac{30}{2*3.142} \\\r= \frac{30}{6.248}\\\r= 4.77 in$

The diameter is 2*r= 4.77*2=9.55 in

Approximately the diameter is 10 inches

7 0

4 years ago

What is the area of an equalateral triangle of side length x?

OLEGan [10]

First you need to make a line that is perpendicular to one side. The diagram is shown below. then the length of the line that is perpendicular to the triangle is
$\frac{ \sqrt{3}x }{2}$
then the area of the equilateral triangle equals to
$\frac{1}{2} height \times base$
so the area would be
$\frac{ \sqrt{3} {x}^{2} }{4}$

5 0

3 years ago

3t - 6 = 18 t = _____

Mazyrski [523]

Answer:

$3t = 18 + 6 \\ 3t = 24 \\ 24 \div 3 = 8 \\ t = 8$

Step-by-step explanation:

First we move the constant to the right and change its sign

we add the numbers and divide it by 3

and you get 8

3 0

3 years ago

Use variation of parameters to find a general solution to the differential equation given that the functions y1 and y2 are linea

Aleks [24]

Recall that variation of parameters is used to solve second-order ODEs of the form

<em>y''(t)</em> + <em>p(t)</em> <em>y'(t)</em> + <em>q(t)</em> <em>y(t)</em> = <em>f(t)</em>

so the first thing you need to do is divide both sides of your equation by <em>t</em> :

<em>y''</em> + (2<em>t</em> - 1)/<em>t</em> <em>y'</em> - 2/<em>t</em> <em>y</em> = 7<em>t</em>

<em />

You're looking for a solution of the form

$y=y_1u_1+y_2u_2$

where

$u_1(t)=\displaystyle-\int\frac{y_2(t)f(t)}{W(y_1,y_2)}\,\mathrm dt$

$u_2(t)=\displaystyle\int\frac{y_1(t)f(t)}{W(y_1,y_2)}\,\mathrm dt$

and <em>W</em> denotes the Wronskian determinant.

Compute the Wronskian:

$W(y_1,y_2) = W\left(2t-1,e^{-2t}\right) = \begin{vmatrix}2t-1&e^{-2t}\\2&-2e^{-2t}\end{vmatrix} = -4te^{-2t}$

Then

$u_1=\displaystyle-\int\frac{7te^{-2t}}{-4te^{-2t}}\,\mathrm dt=\frac74\int\mathrm dt = \frac74t$

$u_2=\displaystyle\int\frac{7t(2t-1)}{-4te^{-2t}}\,\mathrm dt=-\frac74\int(2t-1)e^{2t}\,\mathrm dt=-\frac74(t-1)e^{2t}$

The general solution to the ODE is

$y = C_1(2t-1) + C_2e^{-2t} + \dfrac74t(2t-1) - \dfrac74(t-1)e^{2t}e^{-2t}$

which simplifies somewhat to

$\boxed{y = C_1(2t-1) + C_2e^{-2t} + \dfrac74(2t^2-2t+1)}$

4 0

3 years ago

3/x+1 -1/2 = 1/3x+3 please help solve

mash [69]

$\frac{3}{x+1}^{(6}-\frac{1}{2}^{(3x+3}=\frac{1}{3x+3}^{(2} \\\\ 18-3x-3=2 \\\\ -3x+15=2\\\\ -3x=2-15 \\\\-3x=-13 \\\\ \boxed{x=\frac{13}{3}}$

4 0

3 years ago