Question

Help I don’t understand at all

Answer 1

Answer:

$1)\ \ 4h^2-13h+6\\2)\ \ 7x^3y^2-x^2y+1\\3)\ \ -7n+2\\4)\ \ -8m+4$

Step-by-step explanation:

1.

Simplify the expression by combining like terms. Remember, like terms have the same variable part, to simplify these terms, one performs operations between the coefficients. Please note that a variable with an exponent is not the same as a variable without the exponent. A term with no variable part is referred to as a constant, constants are like terms.

$2h^2-7h+2h^2-h+6+4h-9h$

$(2h^2+2h^2)+(-7h-h+4h-9h)+(6)$

$4h^2-13h+6$

2.

Use a very similar method to solve this problem as used in the first. Please note that all of the rules mentioned in the first problem also apply to this problem; for that matter, the rules mentioned in the first problem generally apply to any pre-algebra problem.

$8x^3y^2-7x^2y+8x-4-x^3y^2+2x^2y+4x^2y-8x+5$

$(8x^3y^2-x^3y^2)+(-7x^2y+2x^2y+4x^2y)+(8x-8x)+(-4+5)$

$7x^3y^2-x^2y+1$

3.

Use the same rules as applied in the first problem. Also, keep the distributive property in mind. In simple terms, the distributive property states the following ($a(b+c)=(a)(b)+(a)(c)=ab+ac$). Also note, a term raised to an exponent is equal to the term times itself the number of times the exponent indicates. In the event that the term raised to an exponent is a constant, one can simplify it. Apply these properties here,

$-2(8n+1)-(5-9n)+3^2$

$-2(8n+1)-(5-9n)+(3*3)$

$-2(8n+1)-(5-9n)+9$

$(-2)(8n)+(-2)(1)+(-)(5)+(-)(-9n)+9$

$-16n-2-5+9n+9$

$(-16n+9n)+(-2-5+9)$

$-7n+2$

4.

The same method used to solve problem (3) can be applied to this problem.

$\frac{1}{2}(10-8m+6m^2)-(3m^2+4m-7)-2^3$

$\frac{1}{2}(10-8m+6m^2)-(3m^2+4m-7)-(2)(2)(2)$

$(\frac{1}{2})(10)+(\frac{1}{2})(-8m)+(\frac{1}{2})(6m^2})+(-)(3m^2)+(-1)(4m)+(-1)(-7)-8$

$5-4m+3m^2-3m^2-4m+7-8$

$(-3m^2-3m^2)+(-4m-4m)+(5+7-8)$

$-8m+4$