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Yanka [14]
3 years ago
13

X+y=4

Mathematics
1 answer:
umka21 [38]3 years ago
7 0

Answer:

3x = 3

Step-by-step explanation:

x + y = 4 × 2

= 2x + 2y = 8

2x + 2y = 8 + x - 2y = -5

3x = 3

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Find the sum of the series<br> {1875 + 750 + 300+...<br> +3.072}
ikadub [295]

Answer:

Find the indicated sum for each geometric series.

{1875+750+300+⋯+3.072}

Answer: 3122.952

Step-by-step explanation:

{1875+750+300+⋯+3.072}

r=0.4 3.072=1875(0.4)n−1

0.0016384=(0.4)n−1

7=n−1→n=8

Sn=1875(1−(0.4)8)1−0.4=3122.952

I think is like this.

7 0
3 years ago
Need a litttle help thx yall
emmasim [6.3K]

Answer:

2 in by 1 in

Step-by-step explanation:

1 5/16 x 2 = 2 5/8

so 2 5/8 = 2 inches

1 5/16 = 1 inch

3 0
3 years ago
1. Which expresses the given inequality in interval notation?
Digiron [165]
1)  x < 4  the answer is  <span>  Option C: x E(-infinity,4)  (it is strict)
2)  </span>2x +6 < 8, and  <span>2x < 8 - 6 = 2, and x<2/2=1
the answer is </span>
<span>D. x E (-infinity, 1)
</span>3)  <span>X greater than or equal to 8 and  x <  - 4
</span>X greater than  8  means x ≥ 8
so we have x ≥ 8 and  <span> x <  - 4
so the answer is  </span>
<span>B. x E (-4,8]
</span> 4) 
7 > x + 6 or x -2 greater than or equal to 3,
<span>7 > x + 6 or x -2 ≥ 3
</span>
<span>7 -6> x, and 1>x
</span>
x -2 ≥ 3   x<span>≥ 3+2=5
finally
</span>x<1 and x≥5   

the answer is
<span>C.x E (1,5]
5)
</span>
|x+3| > 12, 
|x+3| = { -(x+3) if <span>x+3<0 or  </span><span>(x+3)  if </span><span>x+3>0}</span>

so, it is  -(x+3)>12 is equivalent to (x+3)< -12  or (x+3)>12
the answer is
<span>B. x + 3 > 12 or x + 3 <-12</span>






7 0
3 years ago
Read 2 more answers
Solve for t. −3t≥39
lorasvet [3.4K]
-3t ≥39

-3t/-3 ≥39/-3

t≤-13 

is your answer hope this helps

Note: only flip the sign when you are dividing a negative number
4 0
3 years ago
Read 2 more answers
Someone help me please I’ll help you in return
goldfiish [28.3K]

Answer:

wait what do we have to do with this we just have to a line in it?

Step-by-step explanation:

I'll put the answer at the comment section

6 0
3 years ago
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