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2 years ago

Create an equation that has exactly one real solution. 6(3x+4)-x=_x+3

Mathematics

1 answer:

OleMash [197]2 years ago

3 0

Answer:

$x(x+5)=x^2-3x+8$

Step-by-step explanation:

There are infinite cases of equations with exactly one real solution. To create something worth the space, this one should suffice

$x(x+5)=x^2-3x+8$

Operating

$x^2+5x=x^2-3x+8$

Simplifying

$5x=-3x+8$

Joining like terms

$8x=8$

Solving

x=1

Testing the result

$1(1+5)=1^2-3(1)+8=>6=6$

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2 years ago

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FrozenT [24]

Answer:

15 possible combinations

Step-by-step explanation:

Given

$Vegetables = 6$

$Selection = 2$

Required

Determine the possible number of combinations

The question emphasizes on "selection" which means "combination".

So; To answer this question, we apply the following combination formula:

$^nC_r = \frac{n!}{(n-r)r!}$

In this case:

$n = 6$

$r = 2$

The formula becomes:

$^6C_2 = \frac{6!}{(6-2)!2!}$

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$^6C_2 = \frac{30}{2}$

$^6C_2 = 15$

<em>Hence, there are 15 possible combinations</em>

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