How to reduce 10/4 i don’t know how to

2 answers:

5 0

10 and 4 are both divisible by 2. so divide 10 by two and you would get 5 but what you do to the top you do to the bottom. so divide 4 by 2 and you get 5/2 or 2 1/2

vredina [299]3 years ago

3 0

answer: 10/4 reduced is 2.5/1 or 2.5.

work: what we are doing is trying to divide ten into groups of four. how many times does 10 fit into four? well, look at the multiples of four, 4, 8, 12... we can see that 8 fits into ten, but 12 doesn't. that means two whole fours can fit into ten (since 4 x 2 = 8) so we still have a 2 left over from the ten. since 2 is one half of four, we can fit in another .5. so, ten can be fit into 2.5 groups. hope this helps, have a great day!

You might be interested in

Find the missing term. () + (7 − 3i) + (5 + 9i) + 13i = 10 − 5i.

kotegsom [21]

X + (7 - 3i) + (5 + 9i) + 13i = 10 - 5i

Subtract 13i from both sides

x + (7 -3i) + (5 + 9i) = 10 - 18i

Subtract (5 + 9i). MAKE SURE YOU SUBTRACT 9i TOO. In other words, distribute the negative and subtract 5 and 9i at the same time.

x + (7 - 3i) = 5 - 27i

Do the same with (7 - 3i). You'll be adding 3i since -(-3i) = 3i.

x = -2 - 24i

5 0

4 years ago

Read 2 more answers

Explain how to derive a formula for sin(A - B + C)

Akimi4 [234]

$\textit{Sum and Difference Identities} \\\\ sin(\alpha - \beta)=sin(\alpha)cos(\beta)- cos(\alpha)sin(\beta) \\\\ cos(\alpha - \beta)= cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ sin(A-B+C)\implies sin[~(\stackrel{Z}{A-B})~+C]\implies sin(Z+C) \\\\\\ sin(Z)cos(C)+cos(Z)sin(C)\implies sin(A-B)cos(C)+cos(A-B)sin(C) \\\\[-0.35em] ~\dotfill\\\\ \textit{and since we know that}\\\\ sin(A-B)\implies sin(A)cos(B)-cos(A)sin(B)$

$cos(A-B)\implies cos(A)cos(B)+sin(A)sin(B)\\\\ then \\\\[-0.35em] ~\dotfill\\\\\ [sin(A)cos(B)-cos(A)sin(B)]cos(C)+[cos(A)cos(B)+sin(A)sin(B)]sin(C) \\\\\\ ~\hfill \begin{array}{cllll} sin(A)cos(B)cos(C)-cos(A)sin(B)cos(C)\\\\ +\\\\ cos(A)cos(B)sin(C)+sin(A)sin(B)sin(C) \end{array}~\hfill$