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vlabodo [156]
4 years ago
5

Help me please due in 5 minutes

Mathematics
1 answer:
ser-zykov [4K]4 years ago
3 0

Answer:

Step-by-step explanation:

(x - 8)/2 = -5

x - 8 = -10

x = -2

(y + 6)/2 = 10

y + 6 = 20

y = 14

(-2, 14)

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Juliette [100K]

Answer:

-2, -5, 2, then 5

Step-by-step explanation:

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3 years ago
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The table shows the price of two cereal brands and the number of ounces per box. Which is the better price per ounce?
Romashka-Z-Leto [24]

Answer:

A) Cereal brand A

B) There is no table.

Step-by-step explanation:

A) divide each cereal price by the amount of ounces. Brand A is $0.13888 (rounds to $0.14). B is $0.14583 (rounds to $0.15) So, since cereal band A is less cost per ounce, its cheaper

B) This question cannot be answered, since there is no table.

5 0
3 years ago
BRAINLIESTT ASAP! PLEASE HELP ME :)
Yuki888 [10]

Let's say we placed dominoes close together such that they only fell over one way: to the right hand side. Let's call the dominoes A and B.

Domino A is to the left of domino B. If we push over domino A, then it will hit domino B to cause it to fall over as well. We have a mini chain reaction of sorts. The fall of A triggers the fall of B, but not the other way around.

For this problem, we can think of domino A as "Jillian gets a raise" and domino B as "she will buy a new car". The raise causes her to get a new car, but not vice versa.

Since we're told in the last sentence she definitely got the raise, this must mean she will definitely get the new car.

In terms of symbols, the law of detachment is stated as

  • If P, then Q
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Side note: some books might use the term "modus ponens" instead of "law of detachment". They're the same thing.

3 0
3 years ago
3,000 is 10 times as much as
nalin [4]

3,000 is 10 times as much as

3000 = 10 times of what number

We need to find 10 times what number gives us 3000

Let the unknown number be x

3000 = 10 times x

3000 = 10x

To solve for x, we divide by 10 on both sides

\frac{3000}{10}=\frac{10x}{10}

300 = x

So, the unknown number is 300

3000 = 10 times 300

3000 is 10 times as much as 300


4 0
3 years ago
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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
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