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earnstyle [38]
4 years ago
7

Factor the polynomial by its greatest common monomial factor. 3b^5 + 15b^4-18b^7

Mathematics
1 answer:
Nesterboy [21]4 years ago
5 0

Answer:

-3b^4 (6b^3 -b-5)

Step-by-step explanation:

Each term is divisible by 3b^4 so we can factor this out

3b^5 + 15b^4-18b^7

3b^4( b+5 -6b^3)

Rewriting the inside term

3b^4 (-6b^3 +b+5)

We can factor out a negative

-3b^4 (6b^3 -b-5)

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What are the next 3 numbers in this sequence? 1,1,2,4,7,13,24,44
Bogdan [553]
81,149, 274.... adding three numbers from the last would help you

like if you take 7 = 4 + 2 +1

or 13 = 7 + 4 +2

similarly 44+24+13 = 81
              81+44 +24 = 149
              149 + 81 +44 = 274

hope you get it!

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Jet travelling at 500 km/h for 4 hours.
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Answer: jet travels a total of 2000 km in total

Step-by-step explanation:

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3 years ago
Read 2 more answers
A graphing calculator is recommended. A crystal growth furnace is used in research to determine how best to manufacture crystals
Sophie [7]

Answer:

a) w_1 = \frac{-2.156 -\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=-54.896

w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=33.336

And since the power can't be negative then the solution would be w = 33.34 watts.

b) 202= 0.1w^2 +2.156 w +20

0.1w^2 +2.156 w-182=

w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-182)}}{2*0.1}=33.222

204= 0.1w^2 +2.156 w +20

0.1w^2 +2.156 w-184=

w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-184)}}{2*0.1}=33.449

So then the range of voltage would be between 33.22 W and 33.45 W.

c)For this case \epsilon = \pm 1 since that's the tolerance 1C

d) \delta_1 =|33.222-33.336|=0.116

\delta_2 =|33.449-33.336|=0.113

So then we select the smalles value on this case \delta =0.113

e) For this case if we assume a tolerance of \epsilon=\pm 1C for the temperature and a tolerance for the power input \delta =0.113 we see that:

lim_{x \to a} f(x) =L

Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C

Step-by-step explanation:

For this case we have the following function

T(w)= 0.1 w^2 +2.156 w +20

Where T represent the temperature in Celsius and w the power input in watts.

Part a

For this case we need to find the value of w that makes the temperature 203C, so we can set the following equation:

203= 0.1w^2 +2.156 w +20

And we can rewrite the expression like this:

0.1w^2 +2.156 w-183=

And we can solve this using the quadratic formula given by:

w =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}

Where a =0.1, b =2.156 and c=-183. If we replace we got:

w_1 = \frac{-2.156 -\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=-54.896

w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=33.336

And since the power can't be negative then the solution would be w = 33.34 watts.

Part b

For this case we can find the values of w for the temperatures 203-1= 202C and 203+1 = 204 C. And we got this:

202= 0.1w^2 +2.156 w +20

0.1w^2 +2.156 w-182=

w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-182)}}{2*0.1}=33.222

204= 0.1w^2 +2.156 w +20

0.1w^2 +2.156 w-184=

w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-184)}}{2*0.1}=33.449

So then the range of voltage would be between 33.22 W and 33.45 W.

Part c

For this case \epsilon = \pm 1 since that's the tolerance 1C

Part d

For this case we can do this:

\delta_1 =|33.222-33.336|=0.116

\delta_2 =|33.449-33.336|=0.113

So then we select the smallest value on this case \delta =0.113

Part e

For this case if we assume a tolerance of \epsilon=\pm 1C for the temperature and a tolerance for the power input \delta =0.113 we see that:

lim_{x \to a} f(x) =L

Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C

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The area of a rectangle is 0.8 square units. The length is 3.2 units and the width is x units.
Mademuasel [1]
The answer is x=0.25
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3 years ago
You randomly select one card from 52 -card deck fend the probability of selecting an ace or a 7
marta [7]

Answer:

1 in 13 cards

Step-by-step explanation:

4\ 52 simplified is 1/13

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4 years ago
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