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2 years ago

Hi I'm back with more Solving Problems Involving Proportional Relationships because we usually do word problems;

Mathematics

1 answer:

Bingel [31]2 years ago

5 0

Answer:

1) 4 wild cards

2) 51.5 bones, rounded up to 52 bones

3) 72 blue balls.

Step-by-step explanation:

1) For every 18 cards in the game, there is 1 wild card.

For 72 cards in the game, there are 72/18 * 1 = 4 wild cards.

2) For every 4 bones, 1 is in the feet.

For 206 bones, 206/4 * 1 = 51.5 are in the feet. Though, you can't have half a bone, so I'd round it up to 52 bones in the feet.

3) For every 6 + 5 = 11 balls, there are 6 blue and 5 red.

For every 132 balls, there are 132/11 * 6 = 72 blue and 132/11 * 5 = 60 red.

There are 72 blue balls.

You might be interested in

Of 30 students, 1/3 play sports. Of those who play sports, 2/5 play soccer.

lana [24]

Answer: 10 or 12

Step-by-step explanation:

6 0

3 years ago

How can 1/5x − 2 = 1/3x + 8 be set up as a system of equations? a . 5y − 5x = −10. 3y − 3x = 24 b. 5y − 5x = −10. 3y + 3x = 24 c

VMariaS [17]

The answer is d. 5y − x = −10. 3y − x = 24

$\frac{1}{5}x-2= \frac{1}{3}x+8=y$
⇒    $y= \frac{1}{5}x-2$
   $y= \frac{1}{3}x+8$
________________________
$y= \frac{x}{5} -2= \frac{x}{5} - \frac{10}{5} = \frac{x-10}{5}$
$y= \frac{x}{3} +8= \frac{x}{3} + \frac{24}{3} = \frac{x+24}{3}$
________________________
$y= \frac{x-10}{5}$ ⇒    $5y=x-10$
$y= \frac{x+24}{3}$   ⇒    $3y=x+24$
________________________
$5y-x=-10$
$3y-x=24$

5 0

3 years ago

Please help with questions<br> 1d) and 1e)<br><br> Any help is much appreciated thanks

Anestetic [448]

Answer:

1d 6.742 1e 6

Step-by-step explanation: By sketching out the graph you can see the positive x intercept to find 1d

You can also replace h with 25 and isolate t to find 1e

5 0

3 years ago

Which statements are true about the graph of the function f(x) = x2 – 8x + 5? Check all that apply.

statuscvo [17]

Answer:

A, D, E are true

Step-by-step explanation:

You have to complete the square to prove A. Do this by first setting the function equal to 0, then moving the 5 to the other side.

$x^2-8x=-5$

Now we can complete the square. Take half the linear term, square it, and add it to both sides. Our linear term is 8 (from the -8x). Half of 8 is 4, and 4 squared is 16. So we add 16 to both sides.

$(x^2-8x+16)=-5+16$

We will do the addition on the right, no big deal. On the left, however, what we have done in the process of completing the square is to create a perfect square binomial, which gives us the h coordinate of the vertex. We will rewrite with that perfect square on the left and the addition done on the right,

$(x-4)^2=11$

Now we will move the 11 back over, which gives us the k coordinate of the vertex.

$(x-4)^2-11=y$

From this you can see that A is correct.

Also we can see that the vertex of this parabola is (4, -11), which is why B is NOT correct.

The axis of symmetry is also found in the h value. This is, by definition, a positive x-squared parabola (opens upwards), so its axis of symmetry will be an "x = " equation. In the case of this type of parabola, that "x = " will always be equal to the h value. So the axis of symmetry is

x = 4, which is why C is NOT correct, either.

We can find the y-intercept of the function by going back to the standard form of the parabola (NOT the vertex form we found by completing the square) and sub in a 0 for x. When we do that, and then solve for y, we find that when x = 0, y = 5. So the y-intercept is (0, 5).

From this you can see that D is also correct.

To determine if the parabola has real solutions (meaning it will go through the x-axis twice), you can plug it into the quadratic formula to find these values of x. I just plugged the formula into my graphing calculator and graphed it to see that it did, indeed, go through the x-axis twice. Just so you know, the values of x where the function go through are (.6833752, 0) and (7.3166248, 0). That's why you need the quadratic formula to find these values.

7 0

3 years ago

The roots of $3x^2 - 4x + 15 = 0$ are the same as the roots of $x^2 + bx + c = 0,$ for some constants $b$ and $c.$ Find the orde

Akimi4 [234]

we are given

quadratic equation as

$3x^2-4x+15=0$