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Kobotan [32]
4 years ago
5

15 points+ brainliest!! please answer i really need this done!! please i need it asap

Mathematics
2 answers:
Nikolay [14]4 years ago
4 0

Answer:

A

Step-by-step explanation:

You do pythagorean theorem on each one. I did this first one and luckily got it right. 9 squared is 81. 12 squared is 144 and 15 squared is 225. Since 15 is the biggest side, it is the hypotenuse. So you add 81 and 144 to equal 225. The first one is in fact a right triangle.

baherus [9]4 years ago
3 0

Step-by-step explanation:

Since 9^2 + 12^2 = 15^2, it can form a right-angled triangle.

Since 2^6 + 6^2 = 40 and 40 ≠ (√38)^2, it cannot form a right-angled triangle.

Since 3^2 + 9^2 = 90 and 90 ≠ (√91)^2, it cannot form a right-angled triangle.

So the only answer is A.

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Use the Adding Integers manipulative to create a tile model for this situation: What is the sum of (-5) + (+3)?
Karo-lina-s [1.5K]

Answer:

-2

Step-by-step explanation:

(-5) + (+3)

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bixtya [17]

Answer:

(-5,-13)

Step-by-step explanation:

Find the vertex of the parabola y=1/5x^2+2x−8.  

In this equation a=15  and b = 2.  

x= -2/2(1/5) = -2/2/5 = -2/1 * 5/2 = -5

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y=1/5(−5)^2+2(−5)−8

y=5−10−8

y=−13

The vertex is (−5 , −13)

8 0
3 years ago
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Help please. I don't know what a vector is in this context
ale4655 [162]

Answer: The vector relies on how the objects of where the magnitude and directions should lay on.

Step-by-step explanation:

So, I’ll help you with these 4 problems. So, for the vector 2/4, it would 2 units up and 4 units to the right. The second would be 5 units down and 6 units to the right. The third one would be 3 units up and 1 unit to the left. The last one would be 10 units downs and 12 units to the left. A simplified way of thinking of this is to just look at the signs of the number and see where the vector’s magnitude and directions would go in that point of view when you visual the graph or count it in your head.

8 0
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Suppose you have 5 riders and 5 horses, and you want to pair them off so that every rider is assigned one horse (and no horse is
maw [93]

There are 120 ways in which 5  riders and 5 horses can be arranged.

We have,

5 riders and 5 horses,

Now,

We know that,

Now,

Using the arrangement formula of Permutation,

i.e.

The total number of ways ^nN_r = \frac{n!}{(n-r)!},

So,

For n = 5,

And,

r = 5

As we have,

n = r,

So,

Now,

Using the above-mentioned formula of arrangement,

i.e.

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Substituting values,

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^5N_5 = \frac{5!}{(5-5)!}

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So,

There are 120 ways to arrange horses for riders.

Hence we can say that there are 120 ways in which 5  riders and 5 horses can be arranged.

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