There were 42 runners to start a race. In the first half of the race, 1 over 3 of them dropped out. In the second half of the race, 4 over 7 of the remaining runners dropped out. How many runners finished the race?

Answers

Answer 1

Answer: 42 runners total

$(1)/(3)$ of 42 (14) dropped out in first half.

42 - 14 = 28 remaining.

$(4)/(7)$ of 28 (16) dropped out in second half.

28 - 16 = 12 runners finished the race.

Answer 2

Answer: 1/3 of 42 is 14 (42 divided by 3 is 14) so u subtract 14 from 42 which is 28. Then you would divide 28 by 7 which is 4 and then since the fraction is 4/7, u would multiply 4 and 4 to get 16. and finally subtract 28 and 16.

answer is 12 runners

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Which equation is y = 2x² – 8x + 9 rewritten in vertex form? A. y = 2(x – 2)² + 9 B. y = 2(x – 2)² + 5 C. y = 2(x – 2)² + 1 D. y = 2(x – 2)² + 17

Answers

The equation y = 2x² – 8x + 9 can be rewritten as y = 2(x – 2)² + 1.To further simplify this equation, you can factor out the 2 from the front to give y = 2(x – 2)(x – 2) + 1. This makes the equation simpler to manipulate and solve for x.

Now, we need to set the equation equal to zero to find the x-intercepts. When y = 0, we have 0 = 2(x – 2)(x – 2) + 1 which can be rewritten as 0 = (x – 2)(x – 2).

To solve this equation, we need to find the roots, or solutions, of the equation. Using the Quadratic Formula, the roots are going to be x = 2 ± √(2² – 4(2)(1)) / 2(2)

This simplifies as x = 2 ± √2 / 4, so our two x-intercepts are x = 0.5 and x = 3.5.

The graph of y = 2(x – 2)² + 1 will have x-intercepts at (0.5, 0) and (3.5, 0). Therefore, the points of intersection of the graph of y = 2(x – 2)² + 1 and the x-axis are (0.5, 0) and (3.5, 0).

Question 24 Multiple Choice Worth 1 points)(8.01 MC)
Two lines, A and B, are represented by equations given below:
Line A: y = x - 4
Line B: y = 3x + 4
Which of the following shows the solution to the system of equations and explains why?
0 (-3,-5), because the point satisfies one of the equations
0 (-3,-5), because the point lies between the two axes
(-4,-8), because the point satisfies both equations
(-4, -8), because the point does not lie on any axis

Answers

Given:

The system of equations is:

Line A: $y=x-4$

Line B: $y=3x+4$

To find:

The solution of given system of equations.

Solution:

We have,

$y=x-4$ ...(i)

$y=3x+4$ ...(ii)

Equating (i) and (ii), we get

$x-4=3x+4$

$-4-4=3x-x$

$-8=2x$

Divide both sides by 2.

$-4=x$

Substituting $x=-4$ in (i), we get

$y=-4-4$

$y=-8$

The solution of system of equations is (-4,-8).

Now verify the solution by substituting $x=-4, y=-8$ in the given equations.

$-8=-4-4$

$-8=-8$

This statement is true.

Similarly,

$-8=3(-4)+4$

$-8=-12+4$

$-8=-8$

This statement is also true.

Therefore, (-4,-8) is a solution of the given system of equations, because the point satisfies both equations. Hence, the correct option is C.

QUESTION 23 Determine if the given solution is correct. If it is not, find the solution. Is -5 the solution to x+6=5-4? no; x=7 yes no; x=3 no; x=15

Answers

Answer:

yes

Step-by-step explanation:

x+6 = 1

x = -6 + 1

x = -5

Solve the equation 8x-12=5x+15

Answers

8x-12=5x+15
add 12 to both sides

8x=5x+27
subtract 5x from both sides

3x=27
divide by 3

x=9

8x = 5x + 27 Add 12 to both sides

3x = 27 Subtract 5x from both sides

x = 9 Divide both sides by 3

(a-6b)(a+6b) pls answer

Answers

$(a-6b)(a+6b)=a^2-36b^2\n\n (a-b)(a+b)=a^2-b^2$

If triangle RST is congruant to triangle ABC, the measure of angle A equals x^2-8x, and the measure of angle C equals 4x-5, and the measure of angle R equals 5x+30 find the measure of angle C [ only an algebraic solution can receive full credit.]

Answers

In any statement like this one: $\triangle RST \cong \triangle ABC$ you can assume that the points match up in the order that you are given them.
This means that $\angle A \cong \angle R$ .
We know that $m\angle A = x^2-8x$ and $m\angle R=5x+30$ , and because they are congruent we can set the two equal to each other.

$x^2-8x=5x+30$
Let's get everything to one side.
$x^2-13x-30=0$
Let's solve by factoring, since it's easy to do with these whole numbers.
We're looking for two number thats add to -30 and multiply to -13...
These would be -15 and 2.
Since our leading coefficient (_x²) is 1, we can factor straight to (x-15)(x+2).
Here's what it would look like if you went through all the steps anyways, though.
$x^2-15x+2x-30=0$
$x(x-15)+2(x-15)=0$
$(x+2)(x-15)=0$
Any value which causes either factor to equal 0 is a solution.
(The second factor wouldn't matter b/c 0 times anything is still 0)
Therefore x = -2 or 15.
Only one of these is possible, however!
If you use x = -2, you will find that the angle measure 4x-5 is negative, which is impossible. In this case, x must be 15.

Let's find the measure of angle C.
$m\angle C=4x-5\ where\ x=15\nm\angle C=4(15)-5\nm\angle C=60-5\n\boxed{m\angle C = 55\°}$

   ΔRST ≡ ΔABC
      <R = <A
   (5x + 30)° = (x² - 8x)°
      5x + 30 = x² - 8x
-x² + 5x + 30 = x² - x² - 8x
-x² + 5x + 30 = -8x
+ 8x    + 8x
-x² + 13x + 30 = 0
x = -(13) ± √((13)² - 4(-1)(30))
   2(-1)
x = -13 ± √(169 + 120)
      -2
x = -13 ± √(289)
   -2
x = -13 ± 17
   -2
x = -13 + 17 U x = -13 - 17
   -2    -2
x = 4 U x = -30
     -2    -2
x = -2    x = 25

<C = 4x - 5    U <C = 4x - 5
<C = 4(-2) - 5 U <C = 4(25) - 5
<C = -8 - 5       U <C = 100 - 5
<C = -13°       U <C = 95°

or

      ΔRST ≡ ΔABC
      <A + <C = <R
   (x² - 8x)°+ (4x - 5)° = (5x + 30)°
       (x² - 8x) + (4x - 5) = (5x + 30)
   (x² - 8x + 4x - 5) = (5x + 30)
         x² - 4x - 5 = 5x + 30
      - 5x    - 5x
   x² - 9x - 5 = 30
   - 30 - 30
       x² - 9x - 35 = 0
x = -(-9) ± √((-9)² - 4(1)(-35))
       2(1)
x = 9 ± √(81 + 140)
      2
x = 9 ± √(221)
2
x = 9 ± 14.86
      2
x = 9 + 14.86 U x = 9 - 14.86
       2 2
x = 23.86 U x = -4.14
   2 2
x = 11.93 U x = -2.07

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