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max2010maxim [7]
3 years ago
8

What is the sum of the series?

Mathematics
1 answer:
otez555 [7]3 years ago
4 0
Okay, the number above the sigma(4) indicates what value you go to, and what's below the sigma tells you where to start, and the stuff on the right is the expression you're using.

So we start with 1 and go to 4, adding all the values in between.

(2(1))^2 + (2(2))^2 + (2(3))^2 + (2(4))^2

2^2 + 4^2 + 6^2 + 8^2

4 + 16 + 36 + 64

120
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Solve for x. -1+14x 12+17
tino4ka555 [31]

Answer:

x=9

Step-by-step explanation:

These two angles are equal since they are alternate exterior angles

-1 +14x = 12x+17

Subtract 12x from each side

-1 +14x-12x = 12x-12x+17

-1+2x= 17

Add 1  to each side

-1+1+2x = 17+1

2x = 18

2x/2 = 18/2

x =9

8 0
3 years ago
Simplify the imaginary number sqr -75
m_a_m_a [10]

Answer:

5i\sqrt{3}

Step-by-step explanation:

Using the rule of radicals

\sqrt{a} × \sqrt{b} ⇔ \sqrt{ab}

and \sqrt{-1} = i

Given

\sqrt{-75}

= \sqrt{25(3)(-1)}

= \sqrt{25}  × \sqrt{3} × \sqrt{-1}

= 5 × \sqrt{3} × i

= 5i\sqrt{3}

3 0
3 years ago
Jonathan’s age is one-third Belindas age now in 12 years time Belinda will be M years old
nataly862011 [7]
I’m almost positive your answer is B
5 0
3 years ago
Read 2 more answers
X(x+9)=0 <br> x=?<br> What does x have to be for the answer to be 0?
yaroslaw [1]
<em>when we solve the equation it will become
x</em>² + 9x<em>= 0
now here a = 1 , b = 9 and c = 0 
by using quaderatic formula
x = -b +,-</em>√<em>b </em>²<em>- 4ac / 2a     ( 2a is divided by all terms )
by putting values
x = - 9 +,</em><em>- </em>√<em>81 / 2(1)               ( </em>√<em>81 = 9)</em><em>
x = -9 + 9 / 2    and x = -9 -9 /2
x = 0                and x = -18 /2 = -9  
now there are two values for x one is- 9 and other is 0 
so we put -9 as the value of x then answer will be zero..</em>
5 0
3 years ago
Read 2 more answers
Edgar said 3/5 is equivalent to 18/32. Check his work by making a<br> table of equivalent ratios.
Katen [24]

Answer:

In a nutshell, \frac{3}{5} is not equivalent to \frac{18}{32}.

Step-by-step explanation:

Now we proceed to demonstrate that Edgar's statement is false:

1) \frac{3}{5} Given

2) \frac{3}{5} \times \frac{2}{2} Modulative property/Existence of multiplicative inverse

3) \frac{6}{10} \frac{a}{b}\times \frac{c}{d} = \frac{a\cdot b}{c\cdot d}

4) \frac{6}{10}\times \frac{3}{3} Modulative property/Existence of multiplicative inverse

5) \frac{18}{30} \frac{a}{b}\times \frac{c}{d} = \frac{a\cdot b}{c\cdot d}/Result

In a nutshell, \frac{3}{5} is not equivalent to \frac{18}{32}.

4 0
3 years ago
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