Answer:
A = (2p + 9) (2p - 9)
B = (x - 9) (x - 4)
Step-by-step explanation:
For A : Rewrite 4p^2 as (2p)^2.
(2p)^2−81
Rewrite 81 as 9^2.
(2p)^2−9^2
Since both terms are perfect squares, factor using the difference of squares formula, a^2 − b^2 = ( a + b ) ( a − b ) where a = 2p and b = 9 .
(2p + 9) (2p − 9)
For B : Consider the form x^2 + bx + c . Find a pair of integers whose product is c and whose sum is b . In this case, whose product is 36 and whose sum is − 13 .
-9, -4
(x - 9) (x - 4)
I hope this helps.
Elimination because u could multiply the second equation by -3 which would cancel the x and allow you to solve for y and then everntually x but yea elimination is the best method
recall in a perfect square trinomial, the middle term is the tale-tell guy, we know the middle term is the product of 2 and the two other guys without the exponent, so in this case 6x = 2*√x² * √c.
Answer:
<MBK because of the isosceles triangle theorem. If two sides of a triangle are congruent, it is isosceles, so the angles opposite them are also congruent.
Step-by-step explanation:
Answer:
a) = 4.5
b) = 3.3
Step-by-step explanation:
Before solving our problems given to us let us under stand the rule of cube roots
It says ![\sqrt[3]{x \times x \times x} = x](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%20%5Ctimes%20x%20%5Ctimes%20x%7D%20%3D%20x)
-----(A)
Also
![\sqrt[3]{x \times x \times x \times y \times y \times y} = x \times y](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%20%5Ctimes%20x%20%5Ctimes%20x%20%5Ctimes%20y%20%5Ctimes%20y%20%5Ctimes%20y%7D%20%3D%20x%20%5Ctimes%20y)
---(B)
Now let us see each part one by one
a) we have
![\sqrt[3]{64} + \sqrt[3]{0.027} + \sqrt[3]{0.008}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B64%7D%20%2B%20%5Csqrt%5B3%5D%7B0.027%7D%20%2B%20%5Csqrt%5B3%5D%7B0.008%7D)
Now 64 = 4 x 4 x 4
0.027 = 0.3 x 0.3 x 0.3
0.008 = 0.2 x 0.2 x 0.2
substituting these values
![\sqrt[3]{4 \times 4 \times 4} + \sqrt[3]{0.3 \times 0.3 \times 0.3} + \sqrt[3]{0.2 \times 0.2 \times 0.2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B4%20%5Ctimes%204%20%5Ctimes%204%7D%20%2B%20%5Csqrt%5B3%5D%7B0.3%20%5Ctimes%200.3%20%5Ctimes%200.3%7D%20%2B%20%5Csqrt%5B3%5D%7B0.2%20%5Ctimes%200.2%20%5Ctimes%200.2%7D)
Applying Rule A in above
4.5
b) we have ![\sqrt[3]{0.3 \times 0.3 \times 0.3 \times 11 \times 11 \times 11}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B0.3%20%5Ctimes%200.3%20%5Ctimes%200.3%20%5Ctimes%2011%20%5Ctimes%2011%20%5Ctimes%2011%7D)
Applying the B rule in this
3.3
