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3 years ago

A sandwich store charges a $10 delivery fee, and $4.50 for each sandwich. What is the total cost for x sandwiches? *

Mathematics

2 answers:

noname [10]3 years ago

8 0

4.50x + 10. Hope this helps

Alex_Xolod [135]3 years ago

5 0

Answer:

Step-by-step explanation:

4.50x however many she buys just multiply that number to 4.50.

hope this helps

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Which expression is equivalent to 4(5j + 7)?

frozen [14]

Answer: 4(5j+7) = 20j+28

Step-by-step explanation:

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3 years ago

Hello, can you please review attached and help me with the answer ?

Elis [28]

Answer:

(a). P(x) = 8x - 112,000

(b). $80,000

Step-by-step explanation:

(a). P(x) = R(x) - C(x)

P(x) = 50x - 112,000 - 42x

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(b). P(24,000) = 8(24,000) - 112,000 = 80,000

----------------------------------------

(a). P(x) = 8x - 112,000

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2 years ago

Find the difference. write the difference in lower term 8 ⅓-5

sergejj [24]

Given data:

The given expression is 8 ⅓-5.

The given expression can be written as,

$\begin{gathered} 8\frac{1}{3}-5=\frac{25}{3}-5 \\ =\frac{25-15}{3} \\ =\frac{10}{3} \end{gathered}$

Thus, the difference of the given expression is 10/3.

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2 years ago

Opposite reciprocal slope lines or parallel or perpendicular

GarryVolchara [31]

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6 0

3 years ago

1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a

katovenus [111]

1. Let a and b be coefficients such that

$\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}$

Combining the fractions on the right gives

$\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}$

$\implies 1 = (2a+b)x + 3a$

$\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23$

so that

$\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}$

2. a. The given ODE is separable as

$x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}$

Using the result of part (1), integrating both sides gives

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C$

Given that y = 1 when x = 1, we find

$\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)$

so the particular solution to the ODE is

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)$

We can solve this explicitly for y :

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)$

$\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|$

$\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|$

$\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}$

2. b. When x = 9, we get

$y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}$

8 0

2 years ago