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3 years ago

A company has to buy computers and printers. Each computer costs $550 and each printer costs $390. If

Mathematics

2 answers:

kobusy [5.1K]3 years ago

3 0

Answer:

Answer:the company bought 12 computers and 4 printers.

Step-by-step explanation:

Step-by-step explanation:

Let x represent the number of computers that the company bought.

Let y represent the number of printers that the company bought.

The company buys a total of 16 machines. It means that

x + y = 16

Each computer costs $550 and each printer costs $390. If the company spends $8160 for all the computers and printers that was bought, it means that

550x + 390y = 8160 - - - - - - - - - - 1

Substituting x = 16 - y into equation 1, it becomes

550(16 - y) + 390y = 8160

8800 - 550y + 390y = 8160

- 550y + 390y = 8160 - 8800

- 160y = - 640

y = - 640/ - 160

y = 4

Substituting y = 4 into x = 16 - y, it becomes

x = 16 - 4

x = 12

kirza4 [7]3 years ago

3 0

Answer:

The company bought 12 computers and 4 printers.

Step-by-step explanation:

Let x represent the number of computers that the company bought.

Let y represent the number of printers that the company bought.

The company buys a total of 16 machines. It means that

x + y = 16

Each computer costs $550 and each printer costs $390. If the company spends $8160 for all the computers and printers that was bought, it means that

550x + 390y = 8160 - - - - - - - - - - 1

Substituting x = 16 - y into equation 1, it becomes

550(16 - y) + 390y = 8160

8800 - 550y + 390y = 8160

- 550y + 390y = 8160 - 8800

- 160y = - 640

y = - 640/ - 160

y = 4

Substituting y = 4 into x = 16 - y, it becomes

x = 16 - 4

x = 12

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A) Find a particular solution to y" + 2y = e^3 + x^3. b) Find the general solution.

Reptile [31]

Answer:

a.P.I= $\frac{e^{3x}}{11}+\frac{1}{2}(x^3-3x)$

b.G.S= $C_1Cos \sqrt2 x+C_2 Sin\sqrt2 x+\frac{1}{11}e^{3x}+\frac{1}{2}(x^3-3x}$

Step-by-step explanation:

We are given that a linear differential equation

$y''+2y=e^{3x}+x^3$

We have to find the particular solution

P.I= $\frac{e^{3x}}{D^2+2}+\frac{x^3}{D^2+2}$

P.I= $\frac{e^{3x}}{3^2+2}+\frac{1}{2} x^3(1+\frac{D^2}{2})^{-2}$

P.I= $\frac{e^{3x}}{11}+\frac{1-2\frac{D^2}{4}+3\frac{D^4}{16}+...}{2}x^3$

P.I= $\frac{e^{3x}}{11}+\frac{x^3-2\frac{\cdot3\cdot 2x}{4}}{2}+0}$ (higher order terms can be neglected

P.I= $\frac{e^{3x}}{11}+\frac{1}{2}(x^3-3x)$

b.Characteristics equation

$D^2+2=0$

$D=\pm\sqrt2 i$

C.F= $C_1cos \sqrt2x+C_2 sin\sqrt2 x$

G.S=C.F+P.I

G.S= $C_1Cos \sqrt2 x+C_2 Sin\sqrt2 x+\frac{1}{11}e^{3x}+\frac{1}{2}(x^3-3x)$

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3 years ago

Which quadratic equation defines the function that has zeros at − 1/12 and 1/4 ?

Cerrena [4.2K]

$\bf \begin{cases}
x=-\frac{1}{2}\implies &x+\frac{1}{2}=0\\\\
x=\frac{1}{4}\implies &x-\frac{1}{4}=0
\end{cases}
\\\\\\
\left( x+\frac{1}{2} \right)\left( x-\frac{1}{4} \right)=\stackrel{y}{0}\implies \stackrel{FOIL}{x^2+\frac{1}{4}x-\frac{1}{8}}=y$