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poizon [28]
3 years ago
8

J^2k^-3(j^2)^2=j^2k^-3(j^4)

Mathematics
1 answer:
tatiyna3 years ago
5 0
Gaikmann jaiieo nkakshbna kkammm
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Can anyone please help me with this
Monica [59]
Srry but u can’t really see the letters dude .-.
5 0
3 years ago
Factorise the following.​
jok3333 [9.3K]

Answer:

iii. (l + m - n)(l - m + n)

vi. (5a -b) ( 5b - a)

Step-by-step explanation:

iii. l² - (m - n)²

= (l + m - n)(l - m + n)

vi. 4(a+b)² - 9(a-b)²

= (2(a+b))² - (3(a-b))²

= (2(a+b) + 3(a-b)) (2(a+b) - 3(a-b))

= (2a+2b + 3a - 3b) (2a+2b - 3a + 3b)

= (5a -b) ( 5b - a)

7 0
3 years ago
Read 2 more answers
I need help here is the imagen and thank you!!
mojhsa [17]

Answer:

so I don't have a calculator with my but first answer is 9 and second answer is 14 squared plus 14 squared and then you square root it. third answer is 24 squared equals x squared plus x squared.

3 0
3 years ago
Equivalent to 6.4 kilograms
skelet666 [1.2K]

Answer:

14.1096 lb (pounds)

Step-by-step explanation:


3 0
3 years ago
What is the solution set for 12=2(3x - 7)^2 +4
svet-max [94.6K]
12=2(3x-7)^2+4 \ \ \ |-4 \\
8=2(3x-7)^2 \ \ \ |\div 2 \\
4=(3x-7)^2 \\
\sqrt{4}=\sqrt{(3x-7)^2} \\
2=|3x-7| \\
3x-7=2 \ \lor \ 3x-7=-2 \\
3x=2+7 \ \lor \ 3x=-2+7 \\
3x=9 \ \lor \ 3x=5 \\
x=\frac{9}{3} \ \lor \ x=\frac{5}{3} \\
x=3 \lor \ x=1 \frac{2}{3} \\
\boxed{x \in \{1 \frac{2}{3}, 3 \} }
3 0
3 years ago
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