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LiRa [457]
3 years ago
8

There are 12 runners in a race. In how many ways can these runners place first , second, and third?

Mathematics
2 answers:
lisov135 [29]3 years ago
5 0
The correct answer should be B
bonufazy [111]3 years ago
3 0
Answer is C 220. because there are 12 runners
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Suppose we want to choose five objects without replacement from 14 distinct objects if the order of the choices is taken into co
Elanso [62]

Answer:

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3 years ago
Plz help ill mark you brainliest too
algol [13]

Answer: I see you have already marked most of these correctly, the permiter of #4 is 28

Step-by-step explanation:

6: Perimeter: 34 Area: 60

7: 340 / 17 Length of base = 20

8: 221 / 17 Length of base = 13

7 0
3 years ago
17=3(g+3)-g. <br><br> I also need the math
Feliz [49]
Let's solve your equation step-by-step.<span>17=<span><span>3<span>(<span>g+3</span>)</span></span>−g</span></span>
Step 1: Simplify both sides of the equation.<span>17=<span><span>3<span>(<span>g+3</span>)</span></span>−g</span></span><span>
Simplify: (Show steps)</span><span>17=<span><span>2g</span>+9</span></span>
Step 2: Flip the equation.<span><span><span>2g</span>+9</span>=17</span>
Step 3: Subtract 9 from both sides.<span><span><span><span>2g</span>+9</span>−9</span>=<span>17−9</span></span><span><span>2g</span>=8</span>
Step 4: Divide both sides by 2.<span><span><span>2g</span>2</span>=<span>82</span></span><span>g=4</span>
Answer:<span>g=<span>4</span></span>
4 0
3 years ago
Read 2 more answers
Can someone explain how you do fractions like adding them? And what rules they have to find the answer?
frozen [14]

Answer:

Hope this image helps you. (sorry if it's tiny writing.)

Step-by-step explanation:

7 0
2 years ago
Match the parabolas represented by the equations with their vertices. y = x2 + 6x + 8 y = 2x2 + 16x + 28 y = -x2 + 5x + 14 y = -
GaryK [48]

Consider all parabolas:

1.

y = x^2 + 6x + 8,\\y=x^2+6x+9-9+8,\\y=(x^2+6x+9)-1,\\y=(x+3)^2-1.

When x=-3, y=-1, then the point (-3,-1) is vertex of this first parabola.

2.

y = 2x^2 + 16x + 28=2(x^2+8x+14),\\y=2(x^2+8x+16-16+14),\\y=2((x^2+8x+16)-16+14),\\y=2((x+4)^2-2)=2(x+4)^2-4.

When x=-4, y=-4, then the point (-4,-4) is vertex of this second parabola.

3.

y =-x^2 + 5x + 14=-(x^2-5x-14),\\y=-(x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}-14),\\y=-((x^2-5x+\dfrac{25}{4})-\dfrac{25}{4}-14),\\y=-((x-\dfrac{5}{2})^2-\dfrac{81}{4})=-(x-\dfrac{5}{2})^2+\dfrac{81}{4}.

When x=2.5, y=20.25, then the point (2.5,20.25) is vertex of this third parabola.

4.

y =-x^2 + 7x + 7=-(x^2-7x-7),\\y=-(x^2-7x+\dfrac{49}{4}-\dfrac{49}{4}-7),\\y=-((x^2-7x+\dfrac{49}{4})-\dfrac{49}{4}-7),\\y=-((x-\dfrac{7}{2})^2-\dfrac{77}{4})=-(x-\dfrac{7}{2})^2+\dfrac{77}{4}.

When x=3.5, y=19.25, then the point (3.5,19.25) is vertex of this fourth parabola.

5.

y =2x^2 + 7x +5=2(x^2+\dfrac{7}{2}x+\dfrac{5}{2}),\\y=2(x^2+\dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x^2+\dfrac{7}{2}x+\dfrac{49}{16})-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x+\dfrac{7}{4})^2-\dfrac{9}{16})=2(x+\dfrac{7}{4})^2-\dfrac{9}{8}.

When x=-1.75, y=-1.125, then the point (-1.75,-1.125) is vertex of this fifth parabola.

6.

y =-2x^2 + 8x +5=-2(x^2-4x-\dfrac{5}{2}),\\y=-2(x^2-4x+4-4-\dfrac{5}{2}),\\y=-2((x^2-4x+4)-4-\dfrac{5}{2}),\\y=-2((x-2)^2-\dfrac{13}{2})=-2(x-2)^2+13.

When x=2, y=13, then the point (2,13) is vertex of this sixth parabola.

3 0
3 years ago
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