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goblinko [34]
2 years ago
13

Match the multiplication problem on the left with the simplified polynomial on the right.

Mathematics
1 answer:
nata0808 [166]2 years ago
8 0

The match of the multiplication problem with the simplified expressions is as follows:

  • 4x(4x²-x+3) ⇒ 16x³ - 4x² + 12x
  • (8x + 1)(2x -3) ⇒ 16x² - 22x - 3
  • 4x² (4x) ⇒ 16x³
  • (2x + 3) (8x² - 4x + 3) ⇒ 16x³ + 16x² - 6x + 9

<h3>Multiplication of polynomial expressions.</h3>

The process involved in multiplying polynomial expression requires using all the parameters in parentheses to multiply each other.

From the given information:

\mathbf{4x(4x^2-x+3)}

By using distributive law m(a+b+c) = ma + mb + mc

\mathbf{=4x(4x^2)+4x(-x)+4x(3))}

= 16x³ - 4x² + 12x

(8x + 1)(2x -3)

By applying FOIL method: (a+b)(c+d) = ac + ad + bc + bd

= (8x (2x)) + (8x(-3)) + (1 × 2x) + (1 × (-3))

= 16x² - 22x - 3

4x² (4x)

Applying exponent rule; \mathbf{a^b*a^c = a^{b+c}}

\mathbf{= 4x^2 (4x)=  x^2*4^{1+1}x}

\mathbf{=  x^2*4^{2}x}

\mathbf{=  4^{2}x^{2+1}}

\mathbf{=  4^{2}x^{3}}

\mathbf{=  16x^{3}}

(2x + 3) (8x² - 4x + 3)

Using distributive parentheses:

= 2x(8x² - 4x + 3) + 3(8x² - 4x + 3)

= 16x³ + 16x² - 6x + 9

Learn more about polynomial expressions here:

brainly.com/question/2833285

#SPJ1

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Five bells begin to ring together and they ring at intervals of 3, 6, 10, 12 and 15 seconds, respectively. How many times will t
Evgen [1.6K]

Answer:

60 times will they ring together at the same second in one hour excluding the one at the end.

Step-by-step explanation:

Given : Five bells begin to ring together and they ring at intervals of 3, 6, 10, 12 and 15 seconds, respectively.

To find : How many times will they ring together at the same second in one hour excluding the one at the end?

Solution :

First we find the LCM of 3, 6, 10, 12 and 15.

2 | 3  6  10  12  15

2 | 3  3   5   6  15

3 | 3  3   5   3  15

5 | 1    1  5   1    5

  | 1    1   1   1     1

LCM(3, 6, 10, 12,15)=2\times 2\times 3\times 5

LCM(3, 6, 10, 12,15)=60

So, the bells will ring together after every 60 seconds i.e. 1 minutes.

i.e. in 1 minute they rand together 1 time.

We know, 1 hour = 60 minutes

So, in 60 minute they rang together 60 times.

Therefore, 60 times will they ring together at the same second in one hour excluding the one at the end.

6 0
3 years ago
The question is in the picture.
dangina [55]
<h2><u>Part A:</u></h2>

Let's denote no of seats in first row with r1 , second row with r2.....and so on.

r1=5

Since next row will have 10 additional row each time when we move to next row,

So,

r2=5+10=15

r3=15+10=25

<u>Using the terms r1,r2 and r3 , we can find explicit formula</u>

r1=5=5+0=5+0×10=5+(1-1)×10

r2=15=5+10=5+(2-1)×10

r3=25=5+20=5+(3-1)×10

<u>So for nth row,</u>

rn=5+(n-1)×10

Since 5=r1 and 10=common difference (d)

rn=r1+(n-1)d

Since 'a' is a convention term for 1st term,

<h3><u>⇒</u><u>rn=a+(n-1)d</u></h3>

which is an explicit formula to find no of seats in any given row.

<h2><u>Part B:</u></h2>

Using above explicit formula, we can calculate no of seats in 7th row,

r7=5+(7-1)×10

r7=5+(7-1)×10 =5+6×10

r7=5+(7-1)×10 =5+6×10 =65

which is the no of seats in 7th row.

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A taxi service charges $4.50 to pick someone up and then $0.75 per mile during the ride.
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