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mafiozo [28]
3 years ago
9

Find the nth term on the sequence 30,27,24,21

Mathematics
2 answers:
ololo11 [35]3 years ago
6 0
The formula would be n= a1- 3
postnew [5]3 years ago
5 0
30,27,24,21,19,16,13,10,7,4,1,-2,-5,-8,-11,-14,-17,-20,-23,-26
just take away 3 from each number
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I need help solving it
Oxana [17]
7. the last one

I = P R T

I = 15.75 P = 500 R = unknown T = 6

15.75 = 500 (r) (6)

divide the T and I

15.75 ÷ 6 = 500 (r) (6) ÷6

2.62 = 500 (r) *get rid of the six*

divide the P with new answer

2.62 ÷ 500 = 500 ÷ 500 *get rid of 500*

0.00524 = r

move decimal to make it in to a percentage

5.24% = R






6 0
2 years ago
Help me plz! Help me plz!
erik [133]
After converting to common denominator:

8/18, 7/18, 6/18, 5/18

so the next fraction is 4/18, but if you have to simplify it’s 2/9


6 0
3 years ago
What is the expression in radical form?<br><br> (4x3y2)310
sertanlavr [38]

Given:

Consider the given expression is

(4x^3y^2)^{\frac{3}{10}}

To find:

The radical form of given expression.

Solution:

We have,

(4x^3y^2)^{\frac{3}{10}}=(2^2)^{\frac{3}{10}}(x^3)^{\frac{3}{10}}(y^2)^{\frac{3}{10}}

(4x^3y^2)^{\frac{3}{10}}=(2)^{\frac{6}{10}}(x)^{\frac{9}{10}}(y)^{\frac{6}{10}}

(4x^3y^2)^{\frac{3}{10}}=(2)^{\frac{3}{5}}(x)^{\frac{9}{10}}(y)^{\frac{3}{5}}

(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{2^3}\sqrt[10]{x^9}\sqrt[5]{y^3}       [\because x^{\frac{1}{n}}=\sqrt[n]{x}]

(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{8y^3}\sqrt[10]{x^9}       [\because x^{\frac{1}{n}}=\sqrt[n]{x}]

Therefore, the required radical form is \sqrt[5]{8y^3}\sqrt[10]{x^9}.

8 0
3 years ago
What is the area formula for trapezoid
asambeis [7]

Answer:A

=

a

+

b

2

h

Step-by-step explanation:

5 0
3 years ago
Please help to answer this question I’m really confused
igor_vitrenko [27]

Answer:

It would be B!

Step-by-step explanation:

D has an obtuse angle, so it can be eliminated.

C has angles that are not vertical, but rather adjacent. Same goes for A!

Hope this helps!

3 0
2 years ago
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