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2 years ago

Question 7:

Mathematics

1 answer:

Mars2501 [29]2 years ago

5 0

Answer:Graph A.

Step-by-step explanation:

Graph A.

Step-by-step explanation:

We will try to understand the graph plotted for the change in temperature for whole day.

In the morning, temperature starts out around 50°F. As the day goes on, the temperature rises first slowly- Slope of the first line will be less.

Then temperature rises more quickly - Slope of the second line will be more than first line.

It stays constant for an hour- third line will be parallel to the x-axis for one hour gap.

Then the temperature dropped down slowly - Last line of the graph will go down.

By the understanding we made for the graph we find graph A is the answer.

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Price is frequently only way to attract consumers to store when merchandise products are comparable. True False

Helga [31]

Answer: True

Explanation:

Yes, this is true that price is the only way to attract the customers to store when merchandise products are comparable as, price is the main key factor for attracted the customers by the factor like sales and discounts. The main purpose for each retailer is that selling there products with good price and attracted the customers by the price and the quality of the product.

3 0

3 years ago

1/9+29/8 what is the answer? ps: i need to show my work

miskamm [114]

1/9 =   1.11
29/8 = 3.63
   -------
   3.73 = 269/72

or....
1/9 >>> 8/72    (72 is the common denominator) (Multiply the N and D by 8)
29/8 >>> 261/72    (72 is the common denominator) (Multiply the N and D by 9)
   -------------
   269/72

6 0

2 years ago

Read 2 more answers

Find a particular solution to the nonhomogeneous differential equation y??+4y?+5y=?10x+e^(?x).

Firdavs [7]

Answer:

A) Particular solution:

$2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

B) Homogeneous solution:

$y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))$

C) The most general solution is

$y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

Step-by-step explanation:

Given non homogeneous ODE is

$y''+4y'+5y=10x+e^{-x}---(1)$

To find homogeneous solution:

$D^{2}+4D+5=0\\D^{2}+4D+4-4+5=0\\\\(D+2)^{2}=-1\\D+2=\pm iD=-2 \pm i\\y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))---(2)$

To find particular solution:

$y_{p}=Ax+B+Ce^{-x}\\\\y'_{p}=A-Ce^{-x}\\y''_{p}=Ce^{-x}\\$

Substituting $y_{p},y'_{p},y''_{p}$ in (1)

$y''_{p}+4y'_{p}+5y_{p}=10x+e^{-x}\\Ce^{-x}+4(A-Ce^{-x})+5(Ax+B+Ce^{-x})=10x+e^{-x}\\$

Equating the coefficients

$5Ax+2Ce^{-x}+4A+5B=10x+e^{-x}\\5A=10\\A=2\\4A+5B=0\\B=-\frac{4A}{5}B=-\frac{8}{5}2C=1\\C=\frac{1}{2}\\So,\\y_{p}=2x+\frac{1}{2}e^{-x}-\frac{8}{5}---(3)\\$