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malfutka [58]
2 years ago
7

7. A car that was originally worth $29,500 depreciates at a rate of $2,500 per year. Find the

Mathematics
1 answer:
VMariaS [17]2 years ago
4 0
The value is 4.5
hope this help
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La expresion 18+2^3 / 4x2
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The variable, y, varies directly with x. if y = 20 when x = 25, what is y when x = 43? Also please explain so I can understand t
Gwar [14]

Answer:

25

Step-by-step explanation:

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1 year ago
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Snezhnost [94]
I believe it's the first option. Hope this helps.
3 0
3 years ago
Read 2 more answers
Multiply. Check picture.
VashaNatasha [74]

The answer is 3x^4-13x^3-x^2-11x+6.

Solution:

Use algebraic identity: a^m\times a^n=a^{m+n}

For example: x^2\times x=x^{2+1}=x^3

Given expression (x^2-5x+2) and (3x^2+2x+3).

To multiply these equations.

(x^2-5x+2)\times(3x^2+2x+3)

             =x^2(3x^2+2x+3)-5x(3x^2+2x+3)+2(3x^2+2x+3)

             =(3x^4+2x^3+3x^2)+(-15x^3-10x^2-15x)+(6x^2+4x+6)

             =3x^4+2x^3+3x^2-15x^3-10x^2-15x+6x^2+4x+6

Combine like terms together.

             =3x^4+(2x^3-15x^3)+(3x^2-10x^2+6x^2)-15x+4x+6

             =3x^4-13x^3-x^2-11x+6

(x^2-5x+2)\times(3x^2+2x+3)=3x^4-13x^3-x^2-11x+6

Hence the answer is 3x^4-13x^3-x^2-11x+6.

3 0
3 years ago
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