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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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Answer:

12.6

Step-by-step explanation:

Given the function, F(x)=20/4+3e−0.2x

Where e is an exponential function,

To get f(3), we need to substitute x = 3 into the given function

F(3) = 20/4 + 3e - 0.2(3)

f(3) = 5 + 3e - 0.6

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f(3) = 12.554

f(3) = 12.6 to the nearest tenth.

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astraxan [27]

Answer: 4\sqrt{3} .

Step-by-step explanation:

Distance formula : Distance between points (a,b) and (c,d) is given by :-

D=\sqrt{(d-b)^2+(b-a)^2}

Distance between points (-4\sqrt{2},\sqrt{12}) \text{ and }(-\sqrt{32}, 2\sqrt{3}).

D=\sqrt{(2\sqrt{3}-(-\sqrt{12}))^2+(-\sqrt{32}-(-4\sqrt{2}))}\\\\=\sqrt{(2\sqrt{3}+\sqrt{2\times2\times3})^2+(-\sqrt{4\times4\times2}+4\sqrt{2})^2}\\\\=\sqrt{(2\sqrt{3}-\sqrt{2^2\times3})^2+(-\sqrt{4^2\times2}+4\sqrt{2})^2}\\\\=\sqrt{(2\sqrt{3}+2\sqrt{3})^2+(-4\sqrt{2}+4\sqrt{2})^2}\\\\=\sqrt{(4\sqrt{3})^2+0}\\\\=4\sqrt{3}\text{ units}

Hence, the correct option is 4\sqrt{3} .

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4 years ago
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