Add all the number then multiply by 1,000
Answer:
Below.
Step-by-step explanation:
f) (a + b)^3 - 4(a + b)^2
The (a+ b)^2 can be taken out to give:
= (a + b)^2(a + b - 4)
= (a + b)(a + b)(a + b - 4).
g) 3x(x - y) - 6(-x + y)
= 3x( x - y) + 6(x - y)
= (3x + 6)(x - y)
= 3(x + 2)(x - y).
h) (6a - 5b)(c - d) + (3a + 4b)(d - c)
= (6a - 5b)(c - d) + (-3a - 4b)(c - d)
= -(c - d)(6a - 5b)(3a + 4b).
i) -3d(-9a - 2b) + 2c (9a + 2b)
= 3d(9a + 2b) + 2c (9a + 2b)
= 3d(9a + 2b) + 2c (9a + 2b).
= (3d + 2c)(9a + 2b).
j) a^2b^3(2a + 1) - 6ab^2(-1 - 2a)
= a^2b^3(2a + 1) + 6ab^2(2a + 1)
= (2a + 1)( a^2b^3 + 6ab^2)
The GCF of a^2b^3 and 6ab^2 is ab^2, so we have:
(2a + 1)ab^2(ab + 6)
= ab^2(ab + 6)(2a + 1).
Steps to solve:
3x+2y=-4
3x=-4-2y
x=- \frac{4}{3}- \frac{2}{3}y [/tex]
Consider, in ΔRPQ,
RP = R (Radius of larger circle)
PQ = r (radius of smaller circle)
We have to find, RQ, by Pythagoras theorem,
RP² = PQ²+RQ²
R² = r²+RQ²
RQ² = R²-r²
RQ = √(R²-r²
Now, as RQ & QS both are tangents of the smaller circle, their lengths must be equal. so, RS = 2 × RQ
RS = 2√(R²-r²)
Answer:
3√2 and -3√2
Step-by-step explanation:
2x^2-36
2(x^2-36)
2(x+3√2)(x-3√2)
