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Taya2010 [7]
3 years ago
13

Solve for x, l need helpppppppp

Mathematics
2 answers:
adoni [48]3 years ago
8 0

Answer:

17.5

Step-by-step explanation:

90-27=73

4x+3 (reverse the equation)

4/70 - 3

x=17.5

zimovet [89]3 years ago
5 0
4x+3+27=90
4x+30=90
90-30 =60
60/4 =15
X=15
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Natalka [10]

Answer:

\int^1_0\int^3_0\int^9_{3y}\frac{6 cos x^2}{5\sqrt z}dxdydz =\frac{18}{5}(1+\frac{sin2}{2})

Step-by-step explanation:

cosine x²= cos x²

Rule

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Given that,

\int^1_0\int^3_0\int^9_{3y}\frac{6 cos x^2}{5\sqrt z}dxdydz

=\int ^1_0[\int^3_0(\int^9_{3y} \frac{6cos x^2}{5\sqrt z}dz)dy]dz

=\int^1_0[\int^3_0([\frac{6cos x^2 \times \sqrt z}{5\times \frac{1}{2}}]^9_{3y})dy]dx

=\int^1_0[\int^3_0([\frac{12cos x^2 \times( \sqrt 9-\sqrt{3y})}{5}])dy]dx

=\int^1_0[\int^3_0([\frac{12cos x^2 \times( 3-\sqrt{3y})}{5}])dy]dx

=\int^1_0[\frac{12cos x^2 \times( 3y-\frac{\sqrt{3}y^\frac{3}{2}}{\frac{3}{2}})}{5}]^3_0dx

=\int^1_0[\frac{12cos x^2 \times( 3.3-\frac{2\sqrt{3}.3^\frac{3}{2}}{3})}{5}]^3_0dx

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=\frac{18}{5}\int^1_0(1+cos2x)dx

=\frac{18}{5}[(x+\frac{sin2x}{2})]^1_0

=\frac{18}{5}(1+\frac{sin2}{2})

6 0
3 years ago
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Answer:

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Step-by-step explanation:

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