Here we can change the variables to real numbers, just to make things easier for us. For the sake of calculation, let’s say that x=10, y=6, and z=1.

So, y years from now, Yann will be x years old. If we plug in our randomly chosen numbers, we get that Yann will be 10 years old 6 years from now. This means that today, he would be x-y years old, or 10-6, so 4 years old today.
Next we ask how old he was z years ago, or in this case, 1 year ago. We know that if you’re 4 years old now, you were 3 years old last year. In other words, you’re 4-1 years old, or x-y-z years old.

Your answer is x-y-z.

8 0

4 years ago

If (x,y) is a solution to the system of equations shown below, what is the product of the y-coordinates of the solutions? x^2+4y

aliina [53]

Answer:

The product of the y-coordinates of the solutions is equal to 3

Step-by-step explanation:

we have

$x^{2}+4y^{2}=40$ -----> equation A

$x+2y=8$ ------> equation B

Solve by graphing

Remember that the solutions of the system of equations are the intersection point both graphs

using a graphing tool

The solutions are the points (2,3) and (6,1)

see the attached figure

The y-coordinates of the solutions are 3 and 1

therefore

The product of the y-coordinates of the solutions is equal to

(3)(1)=3

4 0

3 years ago

Anthony owns a bakery and buys raw eggs in packs of 50 for $5.00 each. The delivery is a set free of $10.00. If Anthony has $100

yan [13]

Answer:

D- The set of whole numbers were 1<_x<18

Step-by-step explanation:

4 0

3 years ago

2v^2-12 =-12v <br> Would really appreciate it loves❤️

Black_prince [1.1K]

For this case we have the following equation:

$2v ^ 2-12 = -12v$

Rewriting we have:

$2v ^ 2 + 12v-12 = 0$

Dividing by 2 to both sides of the equation:

$v ^ 2 + 6v-6 = 0$

We apply the quadratic formula:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}$

We have to:

$a = 1\\b = 6\\c = -6$

Substituting:

$x = \frac {-6 \pm \sqrt {6 ^ 2-4 (1) (- 6)}} {2 (1)}\\x = \frac {-6 \pm \sqrt {36 + 24}} {2}\\x = \frac {-6 \pm \sqrt {60}} {2}\\x = \frac {-6 \pm \sqrt {4 * 15}} {2}\\x = \frac {-6 \pm2 \sqrt {15}} {2}$

Thus, we have two roots:

$x_ {1} = - 3+ \sqrt {15}\\x_ {2} = - 3- \sqrt {15}$

ANswer:

$x_ {1} = - 3+ \sqrt {15}\\x_ {2} = - 3- \sqrt {15}$

7 0

3 years ago