Hey, i need with these (image attached)

1. (3/5a^3b -5/6ab^3)^2

2. (x/3 +y/5)^2

Step-by-step explanation:

As you know, the square of a binomial expands like this:

(p +q)^2 = p^2 +2pq +q^2

The two "end" terms of the expansion are the squares of the terms of the binomial, and the middle term of the expansion is twice the product of the binomial terms (with the same sign as the binomial).

So, you can work this by ...

• check to see if the "end" terms are perfect squares. If they are, write the binomial as their sum, with the sign of the middle term

• check to see if the middlt term is double the product of the two binomial terms you just wrote. If it is, complete the problem by indicating the square of that binomial.

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1. The square root of 9/25a^6b^2 is 3/5a^3b, so that will be one term of the binomial.

The square root of 25/36a^2b^6 is 5/6ab^3, so that will be the other term of the binomial. The sign of the middle term -a^4b^4 is negative, so our binomial is ...

(3/5a^3b -5/6ab^3)

The product of these two terms is (-1/2)a^4b^4, and the given middle term is twice that. The given trinomial is the square ...

9/25a^6b^2 -a^4b^4 +25/36a^2b^6 = (3/5a^3b -5/6ab^3)^2

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2. The square root of 1/9x^2 is x/3. The square root of 1/25y^2 is y/5. The middle term, 2/15xy, is double the product of these terms, so the expression is the square ...

1/9x^2 +2/15xy +1/25y^2 = (x/3 +y/5)^2

x= -3 and

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

where a is the cube root of .

here   this is 27 and the cube root of 27 is 3. so we write using the form:

.

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

where a = 1, b=3, and c=9

this cubic has two complex roots.

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.