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CaHeK987 [17]
3 years ago
6

A box has a volume given by the trinomial x^3+4x^2-5x. What are possible dimensions of the box? Use factoring.

Mathematics
1 answer:
cricket20 [7]3 years ago
6 0
Answer: The correct answer is Choice B.

To factor this expression, we should first factor out the variable x.

We will have:
x(x^2 + 4x - 5)

Now, all we have to do is factor x^2 + 4x - 5.

The factors of this are (x + 5)(x - 1).

If you put all of the factors together, you have: x(x + 5)(x - 1) or Choice B.
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Please help 100 points.​
IceJOKER [234]

Answer:

76 square feet

Step-by-step explanation:

surface area = 2(wl + hl + hw)      w=width     h=height       l=length

surface area = 2(5*4 + 2*4 + 2*5)

surface area = 2(20 + 8 + 10)

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Write the equation of the line with a slope of 17 and passing through the point (0, 0)
Mkey [24]
The equation for a line is y=mx+b. where m is the slope and b is the y intercept.
The slope of 17 is provided. This means m=17. The point (0,0) tells you the y intercept is 0. This means b=0

Plugging these into the equation and simplifying:
y=17x+0

y=17x

Hope this helps :)
8 0
3 years ago
Find the amount of $8000 for 3 years,compounded annually at 5% per annum. Also ,find the compound interest
Triss [41]

Answer:

$9261

$1261

Step-by-step explanation:

Principal: $8000

Interest rate: 5% PA compounded annually

Time: 3 years

  • Sum = $8000*(1.05)³ = $9261
  • Interest = $9261 - $8000 = $1261
3 0
3 years ago
Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.
frutty [35]
<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

2^n\leq 2^{n+1}-2^{n-1}-1

where n is a positive integer.

  • Let us take n=1

then we have:

2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2

Hence, the result is true for n=1.

  • Let us assume that the result is true for n=k

i.e.

2^k\leq 2^{k+1}-2^{k-1}-1

  • Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u>  2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Let us take n=k+1

Hence, we have:

2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)

( Since, the result was true for n=k )

Hence, we have:

2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2

Also, we know that:

-2

(

Since, for n=k+1 being a positive integer we have:

2^{(k+1)+1}-2^{(k+1)-1}>0  )

Hence, we have finally,

2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

2^n\leq 2^{n+1}-2^{n-1}-1  where n is a positive integer.

6 0
3 years ago
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