Find the value of the following expression: (28 ⋅ 5−5 ⋅ 190)−2 ⋅ 5 to the power of negative 2 over 2 to the power of 3, whole to the power of 4 ⋅ 228 (5 points) Write your answer in simplified form. Show all of your steps. (5 points)
use PEMDAS
can you explain the answer to me please
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y u deleted ur comments
i didnt
yea u did
srr my computer keeps freezing
\[(28 \times 5-5 \times 190)^{-2}\]
\[(\frac{ 5^{-2} }{ 2^{3} })^{4}\]
so can you tell me what the anser for part A and part B will be and plexplain please
what is 228?
can you jus tell me the answer and explain them please and i will medal you
\[(28 \times 5−5 \times 190)^{-2} \times (\frac{ 5^{-3}}{ 2^{3} })^{4} \times 228 \]
i am not here for medal
now first u solve what ever is in bracket
multiply first then subtract can u plz do that
i am refering to the first bracket
\[(-810)^{-2} \times (\frac{ 1 }{ 5^{2} \times 2^{3} })^{4} \times 228\]
\[(-810)^{-2} \times (\frac{ 1^{4} }{ 200^{4}}) \times 228\]
so what the aswer for part A?
\[-\frac{ 1 }{ 810^{2} } \times \frac{ 1 }{ 200^{4} } \times 228\]
where is the part A
there is no part A in the question just say solve
O YEA SORRY
LOL whats the answer for this question can you please explain
\[-\frac{ 1 }{ 656100} \times \frac{ 1 }{ 1600000000 } \times 228\]
cant u seem to follow the steps i am doing
u need to write all these steps to get 5 points
i need to hurry up and get the answer because this has a time limit i have 5 more minutes thats y i just asked for the answer so please can you give me the answer and explain it pplease
p Parentheses first E Exponents (ie Powers and Square Roots, etc.) MD Multiplication and Division (left-to-right) AS Addition and Subtraction (left-to-right)
use a calculator and calculate , i have given u everything
@imqwerty does my solution makes sense?
well what u did is all correct but usually we get nice solutions nd not like that we r getting here maybe the question is not interpreted well... but if what u wrote is the question then the solution is correct :)
@imqwerty yea i was wondering the same thing
:)
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