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maw [93]
2 years ago
15

What is the sign of the point pf the product of nineteen negatice numbers?

Mathematics
1 answer:
Masteriza [31]2 years ago
6 0
A. Negative ♡

Im pretty sure about this

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How do I solve this
pashok25 [27]

Answer: Buses 385

Vans 25

4 0
2 years ago
A sporting goods store sold 2.5 times as many footballs as basketballs last year. Which statement is true? The ratio of football
Mrac [35]
Footballs : basketballs = 2.5 : 1
(x 2)
footballs : basketballs = 5 : 2
⇒ basketballs : footballs = 2 : 5

Answer: <span>The ratio of basketballs to footballs is 5 : 2.</span>
7 0
2 years ago
Prove for any positive integer n, n^3 +11n is a multiple of 6
suter [353]

There are probably other ways to approach this, but I'll focus on a proof by induction.

The base case is that n = 1. Plugging this into the expression gets us

n^3+11n = 1^3+11(1) = 1+11 = 12

which is a multiple of 6. So that takes care of the base case.

----------------------------------

Now for the inductive step, which is often a tricky thing to grasp if you're not used to it. I recommend keeping at practice to get better familiar with these types of proofs.

The idea is this: assume that k^3+11k is a multiple of 6 for some integer k > 1

Based on that assumption, we need to prove that (k+1)^3+11(k+1) is also a multiple of 6. Note how I've replaced every k with k+1. This is the next value up after k.

If we can show that the (k+1)th case works, based on the assumption, then we've effectively wrapped up the inductive proof. Think of it like a chain of dominoes. One knocks over the other to take care of every case (aka every positive integer n)

-----------------------------------

Let's do a bit of algebra to say

(k+1)^3+11(k+1)

(k^3+3k^2+3k+1) + 11(k+1)

k^3+3k^2+3k+1+11k+11

(k^3+11k) + (3k^2+3k+12)

(k^3+11k) + 3(k^2+k+4)

At this point, we have the k^3+11k as the first group while we have 3(k^2+k+4) as the second group. We already know that k^3+11k is a multiple of 6, so we don't need to worry about it. We just need to show that 3(k^2+k+4) is also a multiple of 6. This means we need to show k^2+k+4 is a multiple of 2, i.e. it's even.

------------------------------------

If k is even, then k = 2m for some integer m

That means k^2+k+4 = (2m)^2+(2m)+4 = 4m^2+2m+4 = 2(m^2+m+2)

We can see that if k is even, then k^2+k+4 is also even.

If k is odd, then k = 2m+1 and

k^2+k+4 = (2m+1)^2+(2m+1)+4 = 4m^2+4m+1+2m+1+4 = 2(2m^2+3m+3)

That shows k^2+k+4 is even when k is odd.

-------------------------------------

In short, the last section shows that k^2+k+4 is always even for any integer

That then points to 3(k^2+k+4) being a multiple of 6

Which then further points to (k^3+11k) + 3(k^2+k+4) being a multiple of 6

It's a lot of work, but we've shown that (k+1)^3+11(k+1) is a multiple of 6 based on the assumption that k^3+11k is a multiple of 6.

This concludes the inductive step and overall the proof is done by this point.

6 0
3 years ago
Read 2 more answers
The safety instructions for a 20 foot ladder say the ladder should not be inclined more than 70 degrees with the ground. suppose
Romashka [77]

Answer: the distance of the base of the house to the foot of the ladder is 6.84 feet

Step-by-step explanation:

The scenario is shown in the attached photo.

Right angle triangle ABC is formed when the ladder leans against the wall of the house.

AC = the height of the ladder

AB = x feet = distance of the base of the house to the foot of the ladder

BC is the wall of the building.

To determine x, we will apply trigonometric ratio

Cos # = adjacent/hypotenuse

Where

# = 70 degrees

Hypotenuse = 20

Adjacent = x

Cos 70 = x/20

x = 20cos70

x = 20 × 0.3420

x = 6.84 feets

4 0
3 years ago
Need answer quick. No need for big explanantion
Leona [35]

Answer:

Step-by-step explanation:

4.30/40=$0.1075

Hope this helps :)

5 0
3 years ago
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