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Maru [420]
3 years ago
11

Make arguments. Sue said that 200+20+23 is the same as 200+30+3. Is she correct? Explain.

Mathematics
1 answer:
Leya [2.2K]3 years ago
4 0
No she is not correct if she said 200+20+23 was the same as 200+40+3 she would have been correct.
You might be interested in
Which Inequality reflects the verbal sentence: "The sum of a and b is at most 50!"
aev [14]
A+b is less than or equal to 50
6 0
3 years ago
Please answer! Correct earns brainliest!
kati45 [8]

Answer:

the answer is 96

Step-by-step explanation:

hope this helps :)

3 0
2 years ago
Read 2 more answers
Mops
Ipatiy [6.2K]

Answer:

270

Step-by-step explanation:

For any arithmetic sequence

nth term is given by

nth term = a + (n-1)d

where a is first term,

d is common difference

d is given by nth term - (n-1)th term

sum of n terms given by

sum = n/2(2a + (n-1)d)

________________________________________________

Given arithmetic sequence

-2,0,2,4,6,8...

first term a = -2

lets take third term as nth term and second term as (n-1)th term to find common difference d.

d = 2 - 0 = 2

using a = -2 , d = 2, n = 18

thus, sum of first 18 terms = n/2(2a + (n-1)d)

                                           =18/2( 2*(-2) + (18-1) 2)

                                           =9 ( -4 + 34)

                                           =9 ( 30) = 270

Thus, sum of first 18 terms is 270.

7 0
3 years ago
Simplify the expression. Assume that all variables represent nonzero real numbers.StartFraction (4 n Superscript 4 Baseline q Su
Alja [10]

Answer:

\frac{ - 3}{ 256  {q}^{10} {n}^{8}  }

Step by step explanation:

\frac{ {(4 {n}^{4} {q}^{5})}^{2}  {(8 {n}^{4} q)}^{-2} }{  {(- 3 {nq}^{9})}^{ - 1}   {(4 {n}^{3} {q}^{9})  }^{3} }

first we will change the terms with negative superscrips to the other side of the fraction

\frac{{(4 {n}^{4} {q}^{5})}^{2}{(- 3 {nq}^{9})}^{ 1}}{{(4 {n}^{3} {q}^{9})}^{3} {(8 {n}^{4} q)}^{2} }

then we will distribute the superscripts

\frac{ {4}^{2} {n}^{2 \times 4} {q}^{2 \times 5} (- 3) {nq}^{9}}{ {4 }^{3}{n}^{3 \times 3} {q}^{9 \times 3} {8 }^{2}{n}^{4 \times 2}  {q}^{2} }

\frac{ {4}^{2} {n}^{8} {q}^{10} (- 3) {nq}^{9}}{ {4 }^{3}{n}^{9} {q}^{27} {8 }^{2}{n}^{8}  {q}^{2} }

as when multiplying two powers that have the same base, we can add the exponents and, to divide podes with the same base, we can subtract the exponents

{4}^{2 - 3}  {q}^{10  + 9 - 2 - 27}  {n}^{8 + 1 - 8 - 9}  {8}^{ - 2}  { (- 3)}^{1}

{4}^{ - 1}  {q}^{ - 10}  {n}^{ - 8}  {8}^{ - 2}  { (- 3)}^{1}

then we will change again the terms with negative superscrips to the other side of the fraction

\frac{ - 3}{ 4 \times  {8}^{2}  {q}^{10} {n}^{8}  }

\frac{ - 3}{ 256  {q}^{10} {n}^{8}  }

4 0
3 years ago
3x + 5y = 78
zysi [14]
Hello here is a solution :
<span>3x + 5y = 78 .... (1)
2x - y = 0 ... (2)
from (2) : y = 2x
in (1) : 3x+5(2x) =78
            13x = 78
x= 6
 but : y =2x    y=2(6) = 12
answer : x=6</span>
5 0
3 years ago
Read 2 more answers
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