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3 years ago

A line passes through the point (-3, -3) and has a slope of 1/2 What is the equation of the line?

1 answer:

3 0

Y-y1=m(x-x1)
(x1,y1) is a given point
m=slope
y-(-3)=1/2(x-(-3))
y+3=1/2(x+3) is an equation

2x+3y=6
x+y=4
multiply second equation by -2 and add to first

2x+3y=6
<u>-2x-2y=-8 +</u>
0x+1y=-2

y=-2
sub back

x+y=4
x-2=4
add 2
x=6

x=6
y=-2
(6,-2)

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What is the slope of the line that passes through (-3, 6) and (6, 8)? Type a numerical answer in the space provided. If necessar

shutvik [7]

2/9

Step-by-step explanation:

Slope in calculated as the ratio of change in y values divided by the change in x values

slope, m=Δy/Δx

Δy=8-6=2

Δx=6--3=9

m=2/9

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To calculate slope of a line :brainly.com/question/11267635

Keywords : slope, line, passes

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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.

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3 years ago

Based on the scatterplot below, what is the most likely value for “Pounds of Apples” when “Number of Apples” is 80?

kherson [118]

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4 years ago

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Vesnalui [34]

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4 years ago

Hint: If Angle XYZ (Exterior Angle Y) is 115, then Angle XYW (Interior Angle Y) must be 65 since they are supplementary angles.

RUDIKE [14]

Since all angles in a triangle are equal to 180 and you already have two angles you just have to subtract.
So, 180-(65 + 45)
Simplify = 180- 110
Equals = 70

4 0

2 years ago