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den301095 [7]
3 years ago
8

Jamal has $15.00 to spend at the concession stand. He buys nachos for $7.50, and he wants to purchase some sour straws for $1.50

each. How many sour straws can Jamal purchase with the money that he has?
Mathematics
1 answer:
frosja888 [35]3 years ago
5 0
15-7.50=7.50
7.50/1.50=?
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Valentin [98]

Answer: (a) e ^ -3x (b)e^-3x

Step-by-step explanation:

I suggest the equation is:

d/dx[integral (e^-3t) dt

First we integrate e^-3tdt

Integral(e ^ -3t dt) as shown in attachment and then we differentiate the result as shown in the attachment.

(b) to differentiate the integral let x = t, and substitute into the expression.

Therefore dx = dt

Hence, d/dx[integral (e ^-3x dx)] = e^-3x

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Step-by-step explanation:

x=30° { ins angle inscribed angle Standing on arc are equal }

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3 years ago
The question is in the picture, someone please answer this.
Lemur [1.5K]

Answer:

1) 25x + 15y = 750

2) 25. On a graph it would look like : (25,0)

3) 15.  On a graph it would look like : (15,0)

4) 25x - 750 = 15y or 15y - 750 = 25x

Step-by-step explanation:

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2 years ago
I have 6 feet of licorice and I want to share it with my friends. If each person gets
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Determine whether the set of all linear combinations of the following set of vector in R^3 is a line or a plane or all of R^3.a.
Temka [501]

Answer:

a. Line

b. Plane

c. All of R^3

Step-by-step explanation:

In order to answer this question, we need to study the linear independence between the vectors :

1 - A set of three linearly independent vectors in R^3 generates R^3.

2 - A set of two linearly independent vectors in R^3 generates a plane.

3 - A set of one vector in R^3 generates a line.

The next step to answer this question is to analyze the independence between the vectors of each set. We can do this by putting the vectors into the row of a R^(3x3) matrix. Then, by working out with the matrix we will find how many linearly independent vectors the set has :

a. Let's put the vectors into the rows of a matrix :

\left[\begin{array}{ccc}-2&5&-3\\6&-15&9\\-10&25&-15\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix  ⇒

\left[\begin{array}{ccc}-2&5&-3\\0&0&0\\0&0&0\end{array}\right]

We find that the second vector is a linear combination from the first and the third one (in fact, the second vector is the first vector multiply by -3).

We also find that the third vector is a linear combination from the first and the second one (in fact, the third vector is the first vector multiply by 5).

At the end, we only have one vector in R^3 ⇒ The set of all linear combinations of the set a. is a line in R^3.

b. Again, let's put the vectors into the rows of a matrix :

\left[\begin{array}{ccc}1&2&0\\1&1&1\\4&5&3\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

\left[\begin{array}{ccc}1&1&1\\0&1&-1\\0&0&0\end{array}\right]

We find that there are only two linearly independent vectors in the set so the set of all linear combinations of the set b. is a plane (in fact, the third vector is equivalent to the first vector plus three times the second vector).

c. Finally :

\left[\begin{array}{ccc}0&0&3\\0&1&2\\1&1&0\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

\left[\begin{array}{ccc}1&1&0\\0&1&2\\0&0&3\end{array}\right]

The set is linearly independent so the set of all linear combination of the set c. is all of R^3.

4 0
3 years ago
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