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IrinaVladis [17]
3 years ago
9

How do you solve 13+4x=1-x

Mathematics
2 answers:
mel-nik [20]3 years ago
6 0
13+4x=1-x
4x=-12-x (minus 13 from each side)
5x=-12 (add x to each side)
You divide -12 by 5x
-12÷5x=2\frac{2}{5}⇒2.4
x=2.4
laiz [17]3 years ago
3 0
<span>Simplifying 13 + 4x = 1 + -1x
 Solving 13 + 4x = 1 + -1x
 Solving for variable 'x'.
  Move all terms containing x to the left, all other terms to the right.
  Add 'x' to each side of the equation.13 + 4x + x = 1 + -1x + x
 Combine like terms: 4x + x = 5x 13 + 5x = 1 + -1x + x
 Combine like terms: -1x + x = 0 13 + 5x = 1 + 0 13 + 5x = 1
 Add '-13' to each side of the equation. 13 + -13 + 5x = 1 + -13
 Combine like terms: 13 + -13 = 0   0 + 5x = 1 + -13 5x = 1 + -13
 Combine like terms: 1 + -13 = -12 5x = -12 Divide each side by '5'. x = -2.4 Simplifying x = -2.4</span>
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Holly missed 7 out of 185 school days this year. Approximately what percent of school days was holly in attendance?
MariettaO [177]
It 96%
subtract 7 from 185 that'll equal 178
use the 178 as the numerator for the days she attended
and 185 as the denominator as the days in total.
178/18= x/100 then solve for x and youll get 96 over 100 or 96%

7 0
3 years ago
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Darya [45]
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3 0
3 years ago
Find the area of the figure.
Law Incorporation [45]
(13*8)/2 =52
11*5=55*2=110
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5 0
2 years ago
Can you guys help me please :(
ale4655 [162]

Answer:

720 m²

Step-by-step explanation:

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5 0
2 years ago
Let z = 0.3(cos(31°) + i sin(31°)) and w = 20(cos(18°) + i sin(18°)).
Yuliya22 [10]

Answer:

Option (C) is correct.

Step-by-step explanation:

Given that z = 0.3(cos(31°) + i sin(31°)) and 20(cos(18°) + i sin(18°)).

The product of both the complex number is

zw = [0.3(cos(31°) + i sin(31°))] x [20(cos(18°) + i sin(18°))]

=(0.3x20)[cos(31°) x cos(18°) + cos(31°) x i sin(18°) + i sin(31°) x cos(31°)+ i sin(31°) x i sin(18°)]

Here, (0.3x20)=6, is the magnitude of zw, which is stretching of 0.3 by 20.

zw=6[cos(31°) x cos(18°) + i{cos(31°) x sin(18°) + sin(31°) x cos(31°)}+ i² sin(31°)x sin(18°)]

As i²= -1, So

zw=6[cos(31°) x cos(18°) + i{cos(31°) x sin(18°) + sin(31°) x cos(31°)} - sin(31°)x sin(18°)]

=6[{cos(31°) x cos(18°)- sin(31°)x sin(18°)} + i{cos(31°) x sin(18°) + sin(31°) x cos(31°)}]

As cosAcosB - sin A sin B = cos (A+B) and cosAsinB + sin A sos B = sin (A+B), so

zw=6[cos(31°+18°) + i{sin(31°+18°)]

Note that, initially the argument of z is 31° and after rotation of 18° in counterclockwise direction, the argument od zw is 31°+18°= 49°. i.e

zw=6(cos(49°) + isin(49°)).

So, zw can be determined by stretching z by a factor of 20, then rotating by 18° counterclockwise.

Hence, option (C) is correct.

4 0
3 years ago
Read 2 more answers
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