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seraphim [82]
3 years ago
9

Create a factorable polynomial with a GCF of 2y. Rewrite that polynomial in two other equivalent forms. Explain how each form wa

s created.
I already made my polynomial, 4y^1 + 6y^3
I just don't understand how to get two equivalent forms(please explain if you can)
Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
6 0
4y^1 + 6y^3...so factor out the GCF for another form

2y(2 + 3y^2) <== here is one form

another form would be to multiply ur equation by a multiple of ur GCF 2, such as 4

4(4y^1 + 6y^3) = 16y^1 + 24y^3 <== another form
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what is the slope and y-intercept? write the equation. this is finally the last problem on my homework
IgorLugansk [536]

(y2-y1)/(x2-x1)

3-2=1

4-2=2

slope: 1/2

y-int: 1

y=1/2x+1

4 0
2 years ago
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Find the distance between the points (-6, – 7) and (5. – 7).
QveST [7]

Answer:

11

Step-by-step explanation:

d=√(5-(-6))²+(-7-(-7))²

d=√(11²+0²)

d=11

5 0
3 years ago
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Thanks for the help :)
aliina [53]
Your welcome I’m glad to help out
8 0
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
Solve for t. d=−16t2+4t t=12±41−4d‾‾‾‾‾‾√ t=8±1−4d‾‾‾‾‾‾√ t=18±1−4d√8 t=12±41−4d√2
dangina [55]

Answer:

t=\dfrac{1}{8}\left(1\pm\sqrt{1-4d}\right)

Step-by-step explanation:

Rearranging your quadratic to standard form, you get ...

16t^2 -4t +d = 0

For the quadratic formula, you have ...

a = 16

b = -4

c = d

so the solution is ...

t=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\t=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(16)(d)}}{2(16)}=\dfrac{4\pm 4\sqrt{1-4d}}{32}\\\\t=\dfrac{1}{8}\left(1\pm\sqrt{1-4d}\right)

7 0
3 years ago
Read 2 more answers
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