Answer:
(4+5)+(3+3) because you have eto break up the numbers
9514 1404 393
Answer:
A = 4, B = 3
Step-by-step explanation:
You require that ...
0.30 × A = 0.40 × B
Then ...
B = (3/4)A . . . . . divide by 0.40
Any pair of numbers that satisfies this equation will be possible values of A and B. One such pair is (A, B) = (4, 3).
The value of x is
.
Solution:
Given expression is
.
Switch both sides.
![8-3 \sqrt[5]{x^{3}}=-7](https://tex.z-dn.net/?f=8-3%20%5Csqrt%5B5%5D%7Bx%5E%7B3%7D%7D%3D-7)
Subtract 8 from both side of the equation.
![8-3 \sqrt[5]{x^{3}}-8=-7-8](https://tex.z-dn.net/?f=8-3%20%5Csqrt%5B5%5D%7Bx%5E%7B3%7D%7D-8%3D-7-8)
![-3 \sqrt[5]{x^{3}}=-15](https://tex.z-dn.net/?f=-3%20%5Csqrt%5B5%5D%7Bx%5E%7B3%7D%7D%3D-15)
Divide by –3 on both side of the equation.
![$\frac{-3 \sqrt[5]{x^{3}}}{-3} =\frac{-15}{-3}](https://tex.z-dn.net/?f=%24%5Cfrac%7B-3%20%5Csqrt%5B5%5D%7Bx%5E%7B3%7D%7D%7D%7B-3%7D%20%3D%5Cfrac%7B-15%7D%7B-3%7D)
![\sqrt[5]{x^{3}}=-5](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Bx%5E%7B3%7D%7D%3D-5)
To cancel the cube root, raise the power 5 on both sides.
![(\sqrt[5]{x^{3}})^5=(-5)^5](https://tex.z-dn.net/?f=%28%5Csqrt%5B5%5D%7Bx%5E%7B3%7D%7D%29%5E5%3D%28-5%29%5E5)

To find the value of x, take square root on both sides.
![\sqrt[3]{x^3}=\sqrt[3]{25}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E3%7D%3D%5Csqrt%5B3%5D%7B25%7D)
![x=5\sqrt[3]{25}](https://tex.z-dn.net/?f=x%3D5%5Csqrt%5B3%5D%7B25%7D)
Hence the value of x is
.
Answer:
The trinomial
is not a perfect square trinomial because the third term is not a square.
Step-by-step explanation:
A trinomial is an expression composed of three terms that are joined together by addition or subtraction.
A perfect square is created when a value is multiplied times itself. Thus,
, making the trinomial a² + 2ab + b² a perfect square.
To show that a trinomial is a perfect square you must show that if the first and third terms are squares, figure out what they're squares of. Multiply those things, multiply that product by 2, and then compare your result with the original quadratic's middle term. If you've got a match (ignoring the sign), then you've got a perfect-square trinomial.
The trinomial
is not a perfect square trinomial because the third term is not a square.
A.) xy^7/2z
b.) 4x^4y^8
c.) 10pq^5r^7s