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Aleksandr-060686 [28]
2 years ago
14

Divide : 5 ( 2X+1 ) ( 3X+5 ) ÷ ( 2X+1 )​

Mathematics
2 answers:
lesantik [10]2 years ago
6 0

Answer:

15x+25

Step-by-step explanation:

cancel out (2x+1) you'll be left with 5(3x+5)

when multiply 5 by 3x+5

andrew-mc [135]2 years ago
4 0

Answer:

15x +25

Step-by-step explanation:

2x+1 cancels out 2x+1

then we're left with 5(3x+5)

5*3x=15x

5*5=25

15x+25

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To Mark or decided to purchase a new DVD player on an installment loan DVD player was $295 Tim agreed to pay $31 per month for 1
Verizon [17]

Answer: ⇒ The finance charge = $77


Step-by-step explanation:

Given: The cost price of DVD player= $295

Since, Tim agreed to pay $31 per month for 12 months .

Therefore, the total amount paid by Tim= 31\times12=\$372

The finance charge =\text{Amount paid by Tim - Cost price of DVD}

⇒ The finance charge =\$372-\ $295

⇒ The finance charge =$77

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3 years ago
What is the simplest multicellular organism?
grigory [225]
The simplest multicecullar organism is alga
7 0
3 years ago
Can someone please teach me how to to this<br> Item: $30 Discount: 15%
seraphim [82]

Answer:

$25.5

Step-by-step explanation:

Sure, so the total of the

Item:

30 Dollars ($)

and the discount price is

15 percent (%)

So 30 times 0.15 because 30 is the price in total and 0.15 is the percentage but in a decimal form.

30 times 0.15 equals 4.5

4.5 is the amount that 15 % equals.

30 minus (-) 4.5 because once again the total is 30 and since we have a discount price then we minus (-) the discount price by the total and you get

A total with the discount price as 25.5 Dollars or $25.5

Hope this helps ;)

5 0
3 years ago
What is x in <br> 10(x-2)= -20
MatroZZZ [7]

Answer:

x = 0

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Find a compact form for generating functions of the sequence 1, 8,27,... , k^3
pantera1 [17]

This sequence has generating function

F(x)=\displaystyle\sum_{k\ge0}k^3x^k

(if we include k=0 for a moment)

Recall that for |x|, we have

\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k

Take the derivative to get

\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k

\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k

Take the derivative again:

\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k

\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k

Take the derivative one more time:

\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k

\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k

so we have

\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}

5 0
3 years ago
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