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4 years ago

15

Killian went to the movies with his mother, 2 brothers, and 2 sisters. The theater had choices of 2 different types of soda and

3 different types of candy. If each person ordered a soda and box of candy, is it possible for each person in Killian's family to order a different combination? Be sure to show your work and explain your answer for full credit!!

1 answer:

Assoli18 [71]4 years ago

4 0

yes it is possible for them all to have a different combination

his brother could have soda A and candy A

while the other brother had soda B and candy B

then his sister could have soda C and candy B

then his other sister could have soda C and candy A

while finally killian could have soda B and candy A

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the question in English

Juan has blue cubes with a 55 mm edge and red cubes with a 45 mm edge. He stacks them in two columns, one of each color; he wants the two columns to be the same height. How many cubes does he need, as a minimum, of each color?

Let

x---------> the number of blue cubes

y--------> the number of red cubes

we know that

Juan wants that the two columns to be the same height

so

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the table in the attached figure

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Read 2 more answers

Write the equation of the quadratic function whose graph passes through <img src="https://tex.z-dn.net/?f=%28-3%2C2%29" id="TexF

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Answer:

$f(x)=x^2+3x+2$

Step-by-step explanation:

We want to write the equation of a quadratic whose graph passes through (-3, 2), (-1, 0), and (1, 6).

Remember that the standard quadratic function is given by:

$f(x)=ax^2+bx+c$

Since it passes through the point (-3, 2). This means that when $x=-3$ , $f(x)=f(-3)=2$ . Hence:

$f(-3)=2=a(-3)^2+b(-3)+c$

Simplify:

$2=9a-3b+c$

Perform the same computations for the coordinates (-1, 0) and (1, 6). Therefore:

$0=a(-1)^2+b(-1)+c \\ \\0=a-b+c$

And for (1, 6):

$6=a(1)^2+b(1)+c\\\\ 6=a+b+c$

So, we have a triple system of equations:

$\left\{ \begin{array}{ll} 2=9a-3b+c &\\ 0=a-b+c \\6=a+b+c \end{array} \right.$

We can solve this using elimination.

Notice that the b term in Equation 2 and 3 are opposites. Hence, let's add them together. This yields:

$(0+6)=(a+a)+(-b+b)+(c+c)$

Compute:

$6=2a+2c$

Let's divide both sides by 2:

$3=a+c$

Now, let's eliminate b again but we will use Equation 1 and 2.

Notice that if we multiply Equation 2 by -3, then the b terms will be opposites. So:

$-3(0)=-3(a-b+c)$

Multiply:

$0=-3a+3b-3c$

Add this to Equation 1:

$(0+2)=(9a-3a)+(-3b+3b)+(c-3c)$

Compute:

$2=6a-2c$

Again, we can divide both sides by 2:

$1=3a-c$

So, we know have two equations with only two variables:

$3=a+c\text{ and } 1=3a-c$

We can solve for a using elimination since the c term are opposites of each other. Add the two equations together:

$(3+1)=(a+3a)+(c-c)$

Compute:

$4=4a$

Solve for a:

$a=1$

So, the value of a is 1.

Using either of the two equations, we can now find c. Let's use the first one. Hence:

$3=a+c$

Substitute 1 for a and solve for c:

$\begin{aligned} c+(1)&=3 \\c&=2 \end{aligned}$

So, the value of c is 2.

Finally, using any of the three original equations, solve for b:

We can use Equation 3. Hence:

$6=a+b+c$

Substitute in known values and solve for b:

$6=(1)+b+(2)\\\\6=3+b\\\\b=3$

Therefore, a=1, b=3, and c=2.

Hence, our quadratic function is:

$f(x)=x^2+3x+2$

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