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vovangra [49]
3 years ago
14

If he makes a profit

Mathematics
2 answers:
Svetllana [295]3 years ago
6 0

Answer:

If who makes a profit????

Step-by-step explanation:

Anarel [89]3 years ago
3 0

Answer:

he gets more money then he had to start with

Step-by-step explanation:

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Graph x=4y , x+y=7.0
Marysya12 [62]

The equation of the lines can be plotted on the graph after calculating the coordinates on each line.

<h3>What is a linear equation?</h3>

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.

We have two linear equation:

x = 4y  

x + y = 7.0

To plot the linear equation first we will find the few coordinates to plot on the coordinate plane.

For the equation of line:

x = 4y

x 0 1 2 3 -1 -2 -3

y 0 4 8 12 -4 -8 -12

For the equation of line:

x + y = 7.0

x 0 1 2 3 -1 -2 -3

y 7 6 5 4 8 9 10

Thus, the equation of the lines can be plotted on the graph after calculating the coordinates on each line.

Learn more about the linear equation here:

brainly.com/question/11897796

#SPJ1

4 0
2 years ago
Read 2 more answers
What is another way to write the expression ​ p⋅(10−2) ​ ?
Tomtit [17]

Answer: A

Step-by-step explanation:

I think it is A because p*(10-2). Can = (p*10)-(p*2)

6 0
3 years ago
What is the length of the diagonal (d) is the solid below
elena-14-01-66 [18.8K]
420!!!!!!!!!!!!!!!!!!
8 0
3 years ago
Lim (n/3n-1)^(n-1)<br> n<br> →<br> ∞
n200080 [17]

Looks like the given limit is

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}

With some simple algebra, we can rewrite

\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)

then distribute the limit over the product,

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, <em>e</em> :

\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n

To make our limit resemble this one more closely, make a substitution; replace 9/(<em>n</em> - 9) with 1/<em>m</em>, so that

\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1

From the relation 9<em>m</em> = <em>n</em> - 9, we see that <em>m</em> also approaches infinity as <em>n</em> approaches infinity. So, the second limit is rewritten as

\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}

Now we apply some more properties of multiplication and limits:

\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9

So, the overall limit is indeed 0:

\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}

7 0
3 years ago
Please explain how to solve this (<img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7Ba%5E5%20b%5E3%7D%20" id="TexFormula1" ti
olchik [2.2K]
2 is the base and (\sqrt[4][a^5 b^3]) is the exponent.

the 4 is called an index and it means to fi d the fourth root of the expression under the (\sqrt[][text]) symbol.\\a fourth root is a factor that was multiplied four times. ex: 2*2*2*2=16. the fourth root of 16 is two. \\when you do square roots, you take the number representing the factor out of the radical. anything else that is not a square root stays under. ex: \\([tex]\sqrt[9][text])\\([tex]\sqrt[3*3*3][text]) there is one set of two 3's so\\([tex]3\sqrt[3][text])\\with a fourth root, you look for groups of four. the symbol of that group is placed outside the radical. anything else stays in.\\ ([tex]\sqrt[4][a^5 b^3])

(\sqrt[4][a*a*a*a*a*b*b*b]) there is one group of 4 a's so

(a\sqrt[4][a* b^3])

there really isn't anything else to do to simplify the expression
5 0
3 years ago
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