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Mila [183]
2 years ago
14

Can someone pls help?? i’ll mark brainlist

Mathematics
2 answers:
SVEN [57.7K]2 years ago
5 0

Answer:

D

Step-by-step explanation:

Vaselesa [24]2 years ago
4 0

Answer:

D

Step-by-step explanation:

The line of symmetry is at (-1, 9).

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Y’all I need to ask a question.
Gnesinka [82]

Answer:

A bank normally is the organization which lends money

Step-by-step explanation:

5 0
3 years ago
Find the slope of the line going through the points (-5, 2) and (-1,5).
Karolina [17]
The slope would be 3/4 because 5 - 2 is 3 and -1 - (-5) is 4. Hope this helps!
4 0
3 years ago
Read 2 more answers
X=3m/2-m <br><br>make m the subject of the formula<br>help me plss
wel

Answer:

2X/(3+X)

Step-by-step explanation:

X = 3m/2-m

Cross multiply both sides

X × (2-m)= 3m

2X-Xm= 3m

2X= 3m+Xm

2X= m(3+X)

Divide both sides by the coefficient of m which is (3+X)

m= 2X/(3+X)

8 0
2 years ago
30 circuits is 10% of how many circuits
iVinArrow [24]

30 circuits is 10% of 300 circuits

4 0
3 years ago
Read 2 more answers
How many elements does A contains if it has 64 subsets​
Amanda [17]
<h3>Answer:  6</h3>

===========================================================

Explanation:

Rule: If a set has n elements in it, then it will have 2^n subsets.

For example, there are n = 3 elements in the set {a,b,c}. This means there are 2^n = 2^3 = 8 subsets. The eight subsets are listed below.

  1. {a,b,c} .... any set is a subset of itself
  2. {a,b}
  3. {a,c}
  4. {b,c}
  5. {a}
  6. {b}
  7. {c}
  8. { } ..... the empty set

Subsets 2 through 4 are subsets with exactly 2 elements.  Subsets 5 through 7 are singletons (aka sets with 1 element). The last subset is the empty set which is a subset of any set. You could use the special symbol \varnothing to indicate the empty set.

For more information, check out concepts relating to the power set.

-------------------

The problem is asking what value of n will make 2^n = 64 true.

You could guess-and-check your way to see that 2^n = 64 has the solution n = 6.

Another approach is to follow these steps.

2^n = 64\\\\2^n = 2^6\\\\n = 6

Which is fairly trivial.

Or you can use logarithms to solve for the exponent.

2^n = 64\\\\\text{Log}\left(2^n\right)=\text{Log}\left(64\right)\\\\n*\text{Log}\left(2\right)=\text{Log}\left(64\right)\\\\n=\frac{\text{Log}\left(64\right)}{\text{Log}\left(2\right)}\\\\n\approx\frac{1.80617997398389} {0.30102999566399} \ \text{ ... using base 10 logs}\\\\n\approx5.99999999999983\\\\

Due to rounding error, we don't land exactly on 6 even though we should.

5 0
2 years ago
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