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2 years ago

How do you solve 8x+3c=2200

Mathematics

1 answer:

GREYUIT [131]2 years ago

5 0

8x + 3c = 2200
11xc= 2200
2200÷11
200
200xc

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Does the graph above have an Hamiltonian circuit, Hamiltonian path, or neither? Paths and Circuits Help

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One such path is

1 → 2 → 0 → 4 → 3

which satisfies the requirement that each vertex is visited exactly once.

There is no Hamiltonian circuit, however, since it is impossible for any Hamiltonian path on this graph to visit vertex 0 exactly once.

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How much money will you have if you started with $1250 and put it in an account that earned 6.7% every year for 14 years?

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Answer:

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Step-by-step explanation:

We can use the formula for compound growth to solve this. The formula is:

$F=P(1+r)^t$

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P is the initial amount invested ($1250)

r is the interest rate, in decimal (6.7% is 0.067)

t is the time in years (14, in our case)

<em>Plugging in all the information</em> we have:

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The account will accrue $3098.93 after 14 years.

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Which equation has the solutions x = -3 ± √3i/2 ?

Maurinko [17]

Answer:Answer is option C : [ $x^{2}$ + 3x + 3 ] =0

Note: None of options matches with given question.

instead of "-3" , there should be "- $\frac{3}{2}$ ".

Step-by-step explanation:

Note: None of options matches with given question.

instead of "-3" , there should be " $\frac{3}{2}$ ".

Here, First thing you have to observe the nature of roots.

∴ x = - $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i and x = - $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$

∴ [ x+( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i) ][ x+( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ]=0

∴ [ $x^{2}$ + x( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i)+ x( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i) + ( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i)( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ]=0

∴ [ $x^{2}$ + $\frac{3}{2}$ x + $\frac{\sqrt{3}}{2}$ ix + $\frac{3}{2}$ x - $\frac{\sqrt{3}}{2}$ ix + (3- $\frac{\sqrt{3}}{2}$ i)(3+ $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + ( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i)( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + $\frac{9}{4}$ - ( $\frac{\sqrt{3}}{2}$ i)( $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + $\frac{9}{4}$ - ( $\frac{3}{4}$ ) $i^{2}$ ] =0

∴ [ $x^{2}$ + 3x + $\frac{9}{4}$ + ( $\frac{3}{4}$ ) ] =0

∴ [ $x^{2}$ + 3x + $\frac{12}{4}$ ] =0

∴ [ $x^{2}$ + 3x + 3 ] =0

Thus, Answer is option C : <em>[ $x^{2}$ + 3x + 3 ] =0 </em>

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