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Usimov [2.4K]
3 years ago
8

the professor is decorating his home and wants to buy some plants. he is interested in three types of plants costing $7, $10, an

d $13. his budget is $150 and wants to buy 15 plants. translate the information into a system of two linear equations in three variables. find all solutions of the system. specify at least three of his options.
Mathematics
1 answer:
Luda [366]3 years ago
3 0
X quantity of first plant which cost $7
y quantity of second plant which cost $10
z quantity of third plant which cost $13

{x+y+z=15
{7x+10y+13z=150
There are 8 solutions with spending $150
Five of them are:
1x$7 +  13x$10 + 1x$13
2x$7  + 11x$10  +2x$13
3x$7 +  9x$10  +3x$13
5x$7  +  5x$10 + 5x$13
6x$7 + 3x$10 + 6x$13
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Step-by-step explanation:


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PLEASE HELP DUE NOW!!!<br> WILL MARK BRAINLIEST!!!
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Answer: -10, 10

Step-by-step explanation:

first lets add the two equations. add the x's, add the y's, add the constants on the right.

we get

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so then divide --10 by -1 because we do the opposite operations to both sides

and we get y = 10

now substitute y = 10 into the first equation

-10x - 9(10) = 10

and then we get -10x - 90 = 10

now add 90 to both sides

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2 years ago
Write down in terms of n, an expression for the nth term of the following sequences: 1,8,15,22,29
Crank

Step-by-step explanation:

t1 = 1 = 7*1 - 6

t2 = 8 = 7* 2 - 6

t3 = 15 = 7 * 3 - 6

t4 = 22 = 7 *4 - 6

t5 = 29 = 7 * 5 - 6

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3 years ago
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Answer:

Step-by-step explanation:

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3 years ago
Soap films and bubbles are colorful because the interference conditions depend on the angle of illumination (which we aren't cov
mylen [45]

Answer:

56.39 nm

Step-by-step explanation:

In order to have constructive interference total optical path difference should be an integral number of wavelengths (crest and crest should be interfered). Therefore the constructive interference condition for soap film can be written as,

2t=(m+\frac{1}{2} ).\frac{\lambda}{n}

where λ is the wavelength of light and n is the refractive index of soap film, t is the thickness of the film, and m=0,1,2 ...

Please note that here we include an additional 1/2λ phase shift due to reflection from air-soap interface, because refractive index of latter is higher.

In order to have its longest constructive reflection at the red end (700 nm)

t_1=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_1=\frac{1}{2} .\frac{700}{(2)*(1.33)}\\ \\ t_1=131.58\ nm

Here we take m=0.

Similarly for the constructive reflection at the blue end (400 nm)

t_2=(m+\frac{1}{2} ).\frac{\lambda}{2.n}\\ \\ t_2=\frac{1}{2} .\frac{400}{(2)*(1.33)}\\ \\ t_2=75.19\ nm

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3 years ago
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