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ki77a [65]
2 years ago
10

It’s something with the 30.60.90 thingy that i don’t know how to do someone help ☹️

Mathematics
1 answer:
babunello [35]2 years ago
6 0
Why is there writing everywhere
explanation: hope that helps
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PLEASE HELP!!!
baherus [9]

Answer:

is 26

Step-by-step explanation:

because i said

7 0
3 years ago
What is 5/8 divided by 32
Citrus2011 [14]
First you need to convert 32 into a fraction, so just add 1 as the denominator; 5/8 ÷ 32/1. But you can't divide fractions, so multiply by the reciprocal. 5/8 · 1/32. You should get 5/256 as your answer. I hope that helped!
7 0
3 years ago
Which recursive formula can be used to generate the sequence shown, where f(1) = 5 and n > 1?
saveliy_v [14]

You can try them out and see which one works.

a: f(2) = f(1) +6 = 5+6 = 11 . . . . . . not this one

b: f(1) = f(2) -6 = -1-6 = -7 . . . . . . not this one (5 ≠ -7)

c: f(2) = f(1) - 6 = 5 - 6 = -1 . . . . . this gives the right f(2)

d: f(2 = -6(f(1) = -6(5) = -30 . . . . not this one

_____

The appropriate choice is ...

... f(n +1) = f(n) - 6

— — — — — 

You can also recognize that the next term is 6 less than the current one, so f(n+1) = f(n) - 6, which corresponds to the 3rd selection.

7 0
3 years ago
Read 2 more answers
Dont know how to solve for x can someone help me
vfiekz [6]
  57 + x + x + 130 = 167
  57 + 2x + 130 = 167
    187 + 2x = 167
  - 187          - 187
---------------------------
    2x = -20
  ------  -------
     2       2

x = -10
5 0
3 years ago
Can we obtain a diagonal matrix by multiplying two non-diagonal matrices? give an example
polet [3.4K]
Yes, we can obtain a diagonal matrix by multiplying two non diagonal matrix.

Consider the matrix multiplication below

\left[\begin{array}{cc}a&b\\c&d\end{array}\right]   \left[\begin{array}{cc}e&f\\g&h\end{array}\right] =  \left[\begin{array}{cc}a e+b g&a f+b h\\c e+d g&c f+d h\end{array}\right]

For the product to be a diagonal matrix,

a f + b h = 0 ⇒ a f = -b h
and c e + d g = 0 ⇒ c e = -d g

Consider the following sets of values

a=1, \ \ b=2, \ \ c=3, \ \ d = 4, \ \ e=\frac{1}{3}, \ \ f=-1, \ \ g=-\frac{1}{4}, \ \ h=\frac{1}{2}

The the matrix product becomes:

\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}\frac{1}{3}&-1\\-\frac{1}{4}&\frac{1}{2}\end{array}\right] = \left[\begin{array}{cc}\frac{1}{3}-\frac{1}{2}&-1+1\\1-1&-3+2\end{array}\right]= \left[\begin{array}{cc}-\frac{1}{6}&0\\0&-1\end{array}\right]

Thus, as can be seen we can obtain a diagonal matrix that is a product of non diagonal matrices.
8 0
3 years ago
Read 2 more answers
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