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3 years ago

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Mr. McKay spent $2.60 for a box of crackers and then divided them evenly between the s students in his classroom, Write an expre

ssion for the cost of crackers for each students. Find the cost for s = 20 students.

1 answer:

notka56 [123]3 years ago

5 0

Answer:

You would go $2.60/s

Step-by-step explanation:

then when you have your total number of students you take the total cost and divide that by the number of students that you have to figure out the cost of crackers for each student

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Which expression is equivalent to (5g^4+5g^3-17g^2+6g)-(3g^4+6g^3-7g^2-12)/g+2

ankoles [38]

Option C:

$2 g^{3}-5 g^{2}+6$

Solution:

Given expression:

$$\frac{\left(5 g^{4}+5 g^{3}-17 g^{2}+6 g\right)-\left(3 g^{4}+6 g^{3}-7 g^{2}-12\right)}{g+2}$

To find which expression is equal to the given expression.

$$\frac{\left(5 g^{4}+5 g^{3}-17 g^{2}+6 g\right)-\left(3 g^{4}+6 g^{3}-7 g^{2}-12\right)}{g+2}$

Expand the term $-\left(3 g^{4}+6 g^{3}-7 g^{2}-12\right):-3 g^{4}-6 g^{3}+7 g^{2}+12$

$$=\frac{5 g^{4}+5 g^{3}-17 g^{2}+6 g- 3 g^{4}-6 g^{3}+7 g^{2}+12}{g+2}$

Arrange the like terms together.

$$=\frac{5 g^{4}- 3 g^{4}+5 g^{3}-6 g^{3}-17 g^{2}+7 g^{2}+6 g+12}{g+2}$

$$=\frac{2 g^{4}- g^{3}-10 g^{2}+6 g+12}{g+2}$

Factor the numerator $2 g^{4}-g^{3}-10 g^{2}+6 g+12=(g+2)\left(2 g^{3}-5 g^{2}+6\right)$

$$=\frac{(g+2)\left(2 g^{3}-5 g^{2}+6\right)}{g+2}$

Cancel the common factor g + 2, we get

$=2 g^{3}-5 g^{2}+6$

Hence option C is the correct answer.

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3 years ago

Choose the solution(s) of the following system of equations x2 + y2 = 6 x2 – y = 6

ElenaW [278]

<h2>Answer</h2>

$x=\sqrt{6}, y=0\\ x=-\sqrt{6} , y=0\\x=\sqrt{5} , y=-1\\x=-\sqrt{5} , y=-1$

Or as ordered pairs: $(\sqrt{6} ,0),(-\sqrt{6} ,0),(\sqrt{5} ,-1),(-\sqrt{5} ,-1)$

<h2>Explanation</h2>

Lets solve our system of equations step by step

$x^2+y^2=6$ equation (1)

$x^2-y=6$ equation (2)

1. Solve for $x^2$ in equation (2)

$x^2-y=6$

$x^2=6+y$ equation (3)

2. Replace equation (3) in equation (1) and solve for $y$

$x^2+y^2=6$

$6+y+y^2=6$

$y^2+y=0$

$y(y+1)=0$

$y=0$ or $y=-1$

3. Replace the values of $y$ in equation (3) and solve for $x$

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$x^2=6+y$

$x^2=6$

$x=\sqrt{6}$ or $x=-\sqrt{6}$

- For $y=-1$

$x^2=6+y$

$x^2=6-1$

$x^2=5$

$x=\sqrt{5}$ or $x=-\sqrt{5}$

So, the solutions of our system of equation are:

$x=\sqrt{6}, y=0\\ x=-\sqrt{6} , y=0\\x=\sqrt{5} , y=-1\\x=-\sqrt{5} , y=-1$

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