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3 years ago

3n *power 2 +10n+7 how to factories and pls tell with solution

Mathematics

2 answers:

tankabanditka [31]3 years ago

5 0

Answer:

(n + 1)(3n + 7)

Step-by-step explanation:

3n² + 10n + 7

Consider the factors of the product of the n² term and the constant term which sum to give the coefficient of the n- term.

product = 3 × 7 = 21 and sum = + 10

The factors are + 3 and + 7

Use these factors to split the n- term

3n² + 3n + 7n + 7 ( factor the first/second and third/fourth terms )

3n(n + 1) + 7(n + 1) ← factor out (n + 1) from each term

= (n + 1)(3n + 7) ← in factored form

Dmitriy789 [7]3 years ago

4 0

Answer:

3n^2+10n+7.[multiply 3 and 7=3×7=21.now, put numbers that are equal to 10 when added and 21 when multiplied.so these numbers are.7+3=10 and 7×3=21]

=3n^2+(7+3)n+7

=3n^2+7n+3n+7

=n(3n+7)+1(3n+7)

=(n+1)(3n+7)

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For f(x)=4sin(x2) between x=0 and x=3, find the coordinates of all intercepts, critical points, and inflection points to two dec

telo118 [61]

Answer:

Intercepts:

x = 0, y = 0

x = 1.77, y = 0

x = 2.51, y = 0

Critical points:

x = 1.25, y = 4

x = 2.17 , y = -4

x = 2.8, y = 4

Inflection points:

x = 0.81, y = 2.44

x = 1.81, y = -0.54

x = 2.52, y = 0.27

Step-by-step explanation:

We can find the intercept by setting f(x) = 0

$4sin(x^2) = 0$

$sin(x^2) = 0$

$x^2 = n\pi$ where n = 0, 1, 2,3, 4, 5,...

$x = \sqrt(n\pi)$

Since we are restricting x between 0 and 3 we can stop at n = 2

So the function f(x) intercepts at y = 0 and x:

x = 0

x = 1.77

x = 2.51

The critical points occur at the first derivative = 0

$f^{'}(x) = 4cos(x^2)2x = 8xcos(x^2) = 0$

$xcos(x^2) = 0$

$x = 0$ or

$cos(x^2) = 0$

$x^2 = \frac{\pi}{2} + n\pi$ where n = 0, 1, 2, 3

$x = \sqrt{\pi(n+1/2)}$

Since we are restricting x between 0 and 3 we can stop at n = 2

So our critical points are at

x = 1.25, $y = f(1.25) = 4sin(1.25^2) = 4$

x = 2.17 , $y = f(2.17) = 4sin(2.17^2) = -4$

x = 2.8, $y = f(2.8) = 4sin(2.8^2) = 4$

For the inflection point, we can take the 2nd derivative and set it to 0

$f^[''}(x) = 8(cos(x^2) - xsin(x^2)2x) = 8cos(x^2) - 16x^2sin(x^2) = 0$

$cos(x^2) = 2x^2sin(x^2)$

$tan(x^2) = \frac{1}{2x^2}$

We can solve this numerically to get the inflection points are at

x = 0.81, $y = f(0.81) = 4sin(0.81^2) = 2.44$

x = 1.81, $y = f(1.81) = 4sin(1.81^2) = -0.54$

x = 2.52, $y = f(2.52) = 4sin(2.52^2) = 0.27$

3 0

3 years ago

If f(x)=3x-2 find f-1 (13)

SOVA2 [1]

Answer:

Firstly

f(x)=3x-2=y then

Interchanging x and y we get

x=3y-2

x+2=3y

y=(x+2)/3

f-1(x)=(x+2)/3

f-1(13)=(13+2)/3

f-1(13)=5

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3 years ago

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Serjik [45]

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3 years ago

A pitcher contains 1.75 liters another contains 1800 milliters of apple juice which one has more juice

Strike441 [17]

1liter=1000ml
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1800ml>1750ml

1800ml is bigger

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3 years ago

Read 2 more answers

Can you help me with number 6? <br> Confused abit <br> Please

Sunny_sXe [5.5K]

You can see the three diagram attached. Each link is labeled with the probability: you have probability 1/6 that a six is rolled, and 5/6 that it is not rolled.

To answer the questions, find the path that brings you to the desired outcome, and multiply all the labels you meet.

First question:

To get three sixes, you have to choose the left path at each roll. The probability is always 1/6, so the answer is

$\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{6^3}$

Second question:

To get no sixes, you have to choose the right path at each roll. The probability is always 5/6, so the answer is

$\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^3}{6^3}$

Third question:

To get exactly one six, it can either be the first, second or third roll.

In all cases, you have to choose the left path once and the right path twice: left-right-right mean that you get the six in the first roll, right-left-right means that you get the six in the second roll, right-right-left means that you get the six in the third roll.

In every case, the left turn has probability 1/6, and the right turn has probability 5/6. The probability of each combination is thus

$\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^2}{6^3}$

And since there are three of these combinations, The answer is

$3\frac{5^2}{6^3}$

Fourth question:

Since the question suggests to use what we already achieved, let's do it: having at least one six is the complementary event of having no sixes at all. If an event has probability p, its complementary has probability 1-p. So, since the probability of no sixes is known, the probability of at least one six is

$1 - \frac{5^3}{6^3}$

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3 years ago