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Shalnov [3]
3 years ago
14

Simplify (6b^2)^2 Please hurry !!

Mathematics
2 answers:
nekit [7.7K]3 years ago
5 0

Answer:

36x^4

Step-by-step explanation:

jek_recluse [69]3 years ago
3 0

Answer:

36b^4

Step-by-step explanation:

PEDMAS

 (6b^2)^2

6^2b2x2 would be your next step, you would have to square 6, and multiply

2 by 2 for the exponents.

 36b^4

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What is the value of x? 20,35,60or70
riadik2000 [5.3K]

You are right it is 20.

Hope this helps.

8 0
3 years ago
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Solve quickly please.
lys-0071 [83]

Find the change between the first two sets:

12-9 = 3

14.4 - 10.8 = 3.6

3.6/3 = 1.2

Now find the change between the last two sets:

12 - -3 = 15

14.4 - 3.6 = 10.8

10.8/15 = 0.72

1.2 does not equal 0.72 so the answer is No

4 0
3 years ago
11. You expect to have the given amount in an account with the given terms. Find how much you can withdraw periodically in order
Orlov [11]

This is a question on Present Value of Annuity and we seek the periodic withdrawal.

\begin{gathered} F\text{VA}=\text{PMT(}\frac{(1+\frac{i}{m})^{nm}-1}{\frac{i}{m}}) \\ \text{Therefore, PMT=}\frac{FVA}{(\frac{(1+\frac{i}{m})^{nm}-1}{\frac{i}{m}})} \end{gathered}

where:

FVA = Future value of annuity

PMT = Periodic Withdrawal

i = interest/discount rate

n = no of years

m = no of compundings per interest period

\text{PMT=}\frac{300000}{(\frac{(1+\frac{0.0445}{365})^{22\times365}-1}{\frac{0.0445}{365}})}=22.01

PMT = $22.01

8 0
1 year ago
Find the midpoint of A and B where A has coordinates (2, 3)<br> and B has coordinates (8, 9).
olga55 [171]

Step-by-step explanation:

<h3> Solution,</h3><h3>Given,</h3><h3>Coordinates of A and B are (2,3) and(8,9) respectively.</h3><h3> Let A(2,3)= (x1,y1) and B(8,9)=(x2,y2)</h3><h3>Midpoint of AB(m)=?</h3><h3>Now,using midpoint formula</h3><h3>m= {(x1+x2)/2 ,(y1+y2)/2}</h3><h3>or m= {(2+8)/2,(3+9)/2}</h3><h3>.°. m=(5,6) </h3>
8 0
2 years ago
Read 2 more answers
What is the modulus of |9+40i|?
Talja [164]

Answer:

41

Step-by-step explanation:

We know that complex numbers are a combination of real and imaginary numbers

Real part is x and imaginary part y is multiplied by i, square root of -1

Modulus of x+iy = \sqrt{x^2+y^2}

Here instead of x and y are given 9 and 40

i.e. 9+40i

Hence to find modulus we square the coefficients add them and then find square root

|9+49i| =\sqrt{9^2+40^2} =\sqrt{1681}

By long division method we find that

|9+40i| =41


6 0
3 years ago
Read 2 more answers
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