All categories

2 years ago

Solve x+8y/4=9y for x

Mathematics

2 answers:

nikklg [1K]2 years ago

6 0

Answer:

x = 7y

Step-by-step explanation:

x+8y/4=9y

Multiply each side by 4 to get rid of the fractions

4(x+8y/4)=9y*4

Distribute

4x +8y = 36y

Subtract 8y from each side

4x +8y-8y = 36y-8y

4x = 28y

Divide each side by 4

4x/4 = 28y/4

x = 7y

Eva8 [605]2 years ago

3 0

x+8y/4=9y

Multiply each side by 4

4(x+8y/4)=9y*4

=>4x +8y = 36y

Subtract 8y from each side

=>4x +8y-8y = 36y-8y

=4x = 28y

Divide each side by 4

=>4x/4 = 28y/4

x = 7y

You might be interested in

What is it tell me please

leva [86]

Answer:

19.995

Step-by-step explanation:

7 0

2 years ago

Pls help if pls do most important 15 number 20 number 26 number 28 number 30 number 17 number 25 number 22 number 24 number if u

Tamiku [17]

Answer:

Step-by-step explanation:

15. $\frac{cotx+1}{cotx-1} = \frac{cotx+cotx.tanx}{cotx-cotx.tanx} = \frac{cotx(1+tanx)}{cotx(1-tanx)} = \frac{1+tanx}{1-tanx} \\$

16. $\frac{1+cos\alpha }{sin\alpha } = \frac{sin\alpha }{1-cos\alpha }$

$=> (1+cos\alpha )(1-cos\alpha ) = sin^{2} \alpha \\=> 1-cos^{2}\alpha =sin^{2} \alpha \\=> sin^{2}\alpha +cos^{2}\alpha =1\\$

=> The clause is correct

17. Do the same (16)

18. $\frac{sinx}{1-cosx}+\frac{sinx}{1+cosx} = \frac{sinx(1+cosx) + sinx(1-cosx)}{(1+cosx)(1-cosx)} = \frac{sinx + sinx.cosx + sinx - sinx.cosx}{1-cos^{2}x} = \frac{2sinx}{sin^{2}x }= \frac{2}{sinx} = 2cosecx$

19.

$\frac{sinA}{1+cosA} + \frac{1+cosA}{sinA}=\frac{2}{sinA} \\=> \frac{sinA}{1+cosA} + \frac{1+cosA}{sinA} - \frac{2}{sinA} \\ = 0\\=> \frac{sinA}{1+cosA} + (\frac{1+cosA}{sinA} - \frac{2}{sinA} \\) = 0\\=> \frac{sinA}{1+cosA} + \frac{1+cosA-2}{sinA} = 0\\=> \frac{sinA}{1+cosA} +\frac{cosA-1}{sinA} = 0\\=> \frac{sin^{2} A+(cosA-1)(1+cosA)}{(1+cosA)sinA}=0\\=> \frac{sin^{2} A+cos^{2} A-1}{(1+cosA)sinA}=0\\=> \frac{0}{(1+cosA)sinA} =0\\$