Y=mx+b, the b is your y-intercept so just rearrange your equation
y=(3/7)x+(2/7)
your y-intercept would be b) (2/7)
The zeros for this function are -2, -1 and a double root of 0.
You can find this by first factoring the polynomial on the inside of the parenthesis. Polynomials like this can be factored by looking for two numbers that multiply to the constant (2) and add up to the second coefficient (3). The numbers 2 and 1 satisfy both of those needs and thus can be used as the numbers in a factoring.
x^2(x^2 + 3x + 2)
x^2(x + 2)(x + 1)
Now to find the zeros, we set each part equal to 0. You may want to split the x^2 into two separate x's for this purpose.
(x)(x)(x + 2)(x + 1)
x = 0
x = 0
x + 2 = 0
x = -2
x + 1 = 0
x = -1
first you would multiply 25 x 3, which is 75 and then multiply 60 x 2 which is 120, then add 120 and 75 together, making the answer 195
Answer:
-12p^7q^9
Step-by-step explanation:
(-2/3Pq^4)^2×(-27P^5q)
= ((-2)^2)/3^2 × P^2 q^4-2 × (-27qp^5)
= ((-2)^2)/3^2 × 27p^7q9
= (((-2)^2 x 27p^7q9)/3^2
= -((4×27p^7q^9)/9)
= -(108p^7q^9)/9
= -12p^7q^9
