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Nastasia [14]
3 years ago
7

A company paid $72 for 2 cases of printer paper each case contained 12 packages of paper next month the company's office manager

needs to order 180 packages of the same paper if the price per package does not change would would the total cost of next month's order
Mathematics
1 answer:
Tems11 [23]3 years ago
7 0
To figure out the cost for 180 packages, you must first calculate how many packages they bought for $72. Since each case holds 12 packages, for $72, the company bought 24 packages (12×2). Then you have to divide 24 into 180 to figure out the cost for those 180 packages and get 7.5. To get the cost of the 180 packages, you must multiply 7.5 (the number of 24 packages that go into 180) by 72 (the cost of one group of 24 packages). You then get 540. The company must pay $540 for the 180 packages.
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Please help . I’ll mark you as brainliest if correct.
MariettaO [177]

Answer:

[4, positive infinity]

Step-by-step explanation:

since f(4) exists, the interval starts at 4, and continues to infinity.

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3 years ago
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Tems11 [23]

Answer:

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2 years ago
Use the x-intercept method to find all real solutions of the equation.<br> x^3-8x^2+9x+18=0
meriva

Answer:

a. x=-1,3,\:or\:6

Step-by-step explanation:

The given equation is;

x^3-8x^2+9x+18=0

To solve by the x-intercept method we need to graph the corresponding function using a graphing calculator or software.

The corresponding function is

f(x)=x^3-8x^2+9x+18

The solution to x^3-8x^2+9x+18=0 is where the graph touches the x-axis.

We can see from the graph  that; the x-intercepts are;

(-1,0),(3,0) and (6,0).

Therefore the real solutions are:

x=-1,3,\:or\:6

8 0
3 years ago
Read 2 more answers
1) Use power series to find the series solution to the differential equation y'+2y = 0 PLEASE SHOW ALL YOUR WORK, OR RISK LOSING
iogann1982 [59]

If

y=\displaystyle\sum_{n=0}^\infty a_nx^n

then

y'=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n

The ODE in terms of these series is

\displaystyle\sum_{n=0}^\infty(n+1)a_{n+1}x^n+2\sum_{n=0}^\infty a_nx^n=0

\displaystyle\sum_{n=0}^\infty\bigg(a_{n+1}+2a_n\bigg)x^n=0

\implies\begin{cases}a_0=y(0)\\(n+1)a_{n+1}=-2a_n&\text{for }n\ge0\end{cases}

We can solve the recurrence exactly by substitution:

a_{n+1}=-\dfrac2{n+1}a_n=\dfrac{2^2}{(n+1)n}a_{n-1}=-\dfrac{2^3}{(n+1)n(n-1)}a_{n-2}=\cdots=\dfrac{(-2)^{n+1}}{(n+1)!}a_0

\implies a_n=\dfrac{(-2)^n}{n!}a_0

So the ODE has solution

y(x)=\displaystyle a_0\sum_{n=0}^\infty\frac{(-2x)^n}{n!}

which you may recognize as the power series of the exponential function. Then

\boxed{y(x)=a_0e^{-2x}}

7 0
3 years ago
solve this system of linear equations. Separate the X- and Y- values with a comma. -9x+2y=-16 19x+3y=41​
Serjik [45]

Answer:

(2, 1)

Step-by-step explanation:

The best way to do this to avoid tedious fractions is to use the addition method (sometimes called the elimination method).  We will work to eliminate one of the variables.  Since the y values are smaller, let's work to get rid of those.  That means we have to have a positive and a negative of the same number so they cancel each other out.  We have a 2y and a 3y.  The LCM of those numbers is 6, so we will multiply the first equation by a 3 and the second one by a 2.  BUT they have to cancel out, so one of those multipliers will have to be negative.  I made the 2 negative.  Multiplying in the 3 and the -2:

3(-9x + 2y = -16)--> -27x + 6y = -48

-2(19x + 3y = 41)--> -38x - 6y = -82

Now you can see that the 6y and the -6y cancel each other out, leaving us to do the addition of what's left:

-65x = -130 so

x = 2

Now we will go back to either one of the original equations and sub in a 2 for x to solve for y:

19(2) + 3y = 41 so

38 + 3y = 41 and

3y = 3.  Therefore,

y = 1

The solution set then is (2, 1)

6 0
3 years ago
Read 2 more answers
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