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polet [3.4K]
3 years ago
6

Factor each Polynomial​

Mathematics
2 answers:
Stels [109]3 years ago
8 0

Answer:

2(+2)(+3)

Step-by-step explanation:

Sindrei [870]3 years ago
6 0

Answer:

\huge\boxed{ \sf 2x (x+3)(x+2)}

Step-by-step explanation:

\sf =2x^3+10x^2+12x\\\\Take \ 2x \ common\\\\=2x(x^2+5x+6)\\\\By \ mid-term \ break \ formula\\\\=2x(x^2+3x+2x+6)\\\\=2x[x(x+3)+2(x+3)]\\\\Take \ (x+3) \ common\\\\= 2x (x+3)(x+2)\\\\\rule[225]{225}{2}

Hope this helped!

<h3>~AnonymousHelper1807</h3>
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A child born today will see more than _____________ advertisements in their life.
Ymorist [56]
The answer would be 1 million because if the average person lives to be 70 multiply that by 2000 and you get 51100000 so I would say that total is closer to 1 million then it is to 3 billion
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3 years ago
ASAP what is the value of x
Nataliya [291]
2x+2= 3x-52

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3 years ago
Find the area of the region enclosed by f(x) and the x axis for the given function over the specified u reevaluate f(x)=x^2+3x+4
lord [1]

Answer:

A = 68 unit^2

Step-by-step explanation:

Given:-

The piece-wise function f(x) is defined over an interval as follows:

                       f(x) =  { x^2+3x+4    , x < 3

                       f(x) =  { x^2+3x+4     , x≥3

                        Domain : [ -3 , 4 ]

Find:-

Find the area of the region enclosed by f(x) and the x axis

Solution:-

- The best way to tackle problems relating to piece-wise functions is to solve for each part individually and then combine the results.

- The first portion of function is valid over the interval [ -3 , 3 ]:      

                       f(x) = x^2+3x+4

- The area "A1" bounded by f(x) is given as:

                      A1 = \int\limits^a_b {f(x)} \, dx

Where,  The interval of the function { -3 , 3 ] = [ a , b ]:

                     A1 = \int\limits^a_b {x^2+3x+4} \, dx\\\\A1 = \frac{x^3}{3} + \frac{3x^2}{2} + 4x |\limits_-_3^3 \\\\A1 = \frac{3^3}{3} + \frac{3*3^2}{2} + 4*3 - \frac{-3^3}{3} - \frac{3(-3)^2}{2} - 4(-3)\\\\A1 = 9 + 13.5 +12 + 9-13.5+12\\\\A1 =42 unit^2

- Similarly for the other portion of piece-wise function covering the interval [3 , 4] :

                     f(x) = 4x+10

- The area "A2" bounded by f(x) is given as:

                      A2 = \int\limits^a_b {f(x)} \, dx

Where,  The interval of the function { 3 , 4 ] = [ a , b ]:

                     A2 = \int\limits^a_b {4x+10} \, dx\\\\A2 = 2x^2 + 10x |\limits_3^4 \\\\A2 = 2*(4)^2 + 10*4 - 2*(3)^2 - 10*3\\\\A2 = 32 + 40 - 18-30\\\\A2 =26 unit^2

- The total area "A" bounded by the piece-wise function over the entire domain [ -3 , 4 ] is given:

                     A = A1 + A2

                     A = 42 + 26

                     A = 68 unit^2

8 0
2 years ago
Find the product of all positive integer values of $c$ such that $3x^2+7x+c=0$ has two real roots. I will award a lot of points
Agata [3.3K]

Answer:  24

<u>Step-by-step explanation:</u>

Let's find one solution:

3x² + 7x + c = 0

a=3 b=7  c=c

First, let's find c so that it has REAL ROOTS.

⇒ Discriminant (b² - 4ac) ≥ 0

                         7² - 4(3)c ≥ 0

                         49 - 12c ≥ 0

                               -12c  ≥ -49

                                c\leq\dfrac{-49}{-12}\quad \rightarrow c\leq \dfrac{49}{12}      

Since c must be a positive integer, 1 ≤ c ≤ 4

Example: c = 4

3x² + 7x + 4 = 0

(3x + 4)(x + 1) = 0

x = -4/3, x = -1         Real Roots!

<em>You need to use Quadratic Formula to solve for c = {1, 2, 3}</em>

<em />

Valid solutions for c are: {1, 2, 3, 4)

Their product is: 1 x 2 x 3 x 4 = 24

<em />

7 0
3 years ago
Becky added 273+416 and got 688.Then she checked her answer by adding 688+416.What error did Becky make?Was the sum she found fo
Komok [63]
You were post to -416 and get 273
5 0
3 years ago
Read 2 more answers
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