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3 years ago

Solve the inequalities by graphing. Identify the graph that shows the following equations.

Mathematics

1 answer:

Yuki888 [10]3 years ago

3 0

Y>2x (0,0)—> (0+2, 0+1)
Y<3 Y=3 then shade downward
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Emily had some candy to give her five children.She first took two pieces for her self and then evenly divided the rest among her

kodGreya [7K]

She started with 17 pieces of candy

8 0

3 years ago

Helpppp pleaseeeee and could you put an explenation??? pleaseeee

Illusion [34]

Answer:

Step-by-step explanation:

The angles would be supplementary

45 + 5x + 35 = 180

5x + 80 = 180

5x = 100

x = 20

To find out why they are supplementary, refer to the transversal below

∠3 = ∠7 (corresponding angles)

∠8 = 180 - ∠7 (supplementary angles)

And since ∠7 = ∠3...

∠8 = 180 - ∠3 which is why (in the problem) those two angles are supplementary (adding to 180 degrees)

4 0

3 years ago

Answer pwease lol now

Anettt [7]

Answer:

D) y = 27 · (3)^x

Step-by-step explanation:

Replace x with the x values.

27 · (3)^-2 = 27 · 1/9 = 3

27 · (3)^-1 = 27 · 1/3 = 9

27 · (3)^0 = 27 · 1 = 27

27 · (3)^1 = 27 · 3 = 81

17 · (3)^2 = 27 · 9 = 243

D is your answer.

Hope this helps

5 0

2 years ago

What is the solution to the inequality-2x-3y>9

Sati [7]

Answer:

y < − 2 x /3 − 3

Hope This Helps!

4 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