<span> The product of two perfect squares is a perfect square.
Proof of Existence:
Suppose a = 2^2 , b = 3^2 [ We have to show that the product of a and b is a perfect square.] then
c^2 = (a^2) (b^2)
= (2^2) (3^2)
= (4)9
= 36
and 36 is a perfect square of 6. This is to be shown and this completes the proof</span>
You multiply 55.25 by 0.18. You get 9.945 and then you add it to 55.25. You should get 65.195. Round it to $65.20. That's your answer.
When naming triangles we name them by the letters on each corner so only answers with T, A, and X in them are correct, so TAX, XTA, and AXT
Answer:
(i) The length of AC is 32 units, (ii) The length of BC is 51 units.
Step-by-step explanation:
(i) Let suppose that AB and BC are collinear to each other, that is, that both segments are contained in the same line. Algebraically, it can be translated into this identity:
![AC = AB + BC](https://tex.z-dn.net/?f=AC%20%3D%20AB%20%2B%20BC)
If we know that
and
, then:
![AC = 8 + 24](https://tex.z-dn.net/?f=AC%20%3D%208%20%2B%2024)
![AC = 32](https://tex.z-dn.net/?f=AC%20%3D%2032)
The length of AC is 32 units.
(ii) Let suppose that AB and AC are collinear to each other, that is, that both segments are contained in the same line. Algebraically, it can be translated into this identity:
![AC = AB + BC](https://tex.z-dn.net/?f=AC%20%3D%20AB%20%2B%20BC)
![BC = AC - AB](https://tex.z-dn.net/?f=BC%20%3D%20AC%20-%20AB)
If we know that
and
, then:
![BC = 68-17](https://tex.z-dn.net/?f=BC%20%3D%2068-17)
![BC = 51](https://tex.z-dn.net/?f=BC%20%3D%2051)
The length of BC is 51 units.