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3 years ago

10

Solve 10 and 11 . Show and explain your work.

1 answer:

DaniilM [7]3 years ago

6 0

Answer:

500-75y=x. And. 44+14y=x. No work needed

Step-by-step explanation:

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4 years ago

Read 2 more answers

Find the values of y = c(x) = for x = 0, 0.008, 0.027, 0.064, 0.125, 0.216, 0.343, 0.512, , 0.729, and 1.

kondaur [170]

c(x) is a function, then we will solve the problem for two different functions.

1. If $c(x)=\frac{x}{0.5}+x$

Then

$y=c(x)$

$y=\frac{x}{0.5}+x$

Using each of the values for x we have.

When x=0 the result is: $y=\frac{0}{0.5}+0=0$

When x=0.008 the result is: $y=\frac{0.008}{0.5}+0.008=0.016+0.008=0.024$

When x=0.027 the result is: $y=\frac{0.027}{0.5}+0.027=0.054+0.027=0.081$

When x=0.064 the result is: $y=\frac{0.064}{0.5}+0.064=0.128+0.064=0.192$

When x=0.125 the result is: $y=\frac{0.125}{0.5}+0.125=0.25+0.125=0.375$

When x=0.216 the result is: $y=\frac{0.216}{0.5}+0.216=0.432+0.216=0.648$

When x=0.343 the result is: $y=\frac{0.343}{0.5}+0.343=0.686+0.343=1.029$

When x=0.512 the result is: $y=\frac{0.512}{0.5}+0.512=1.024+0.512=1.536$

When x=0.729 the result is: $y=\frac{0.729}{0.5}+0.729=1.458+0.729=2.187$

When x=1 the result is: $y=\frac{1}{0.5}+1=2+1=3$

2. If $c(x)=sin(x)+2x$

Then

$y=c(x)$

$y=sin(x)+2x$

Using each of the values for x we have.

When x=0 the result is: $y=sin(0)+2(0)=0$

When x=0.008 the result is: $y=sin(0.008)+2(0.008)=0.000139+0.016=0.016139$

When x=0.027 the result is: $y=sin(0.027)+2(0.027)=0.000471+0.054=0.054471$

When x=0.064 the result is: $y=sin(0.064)+2(0.064)=0.001117+0.128=0.129117$

When x=0.125 the result is: $y=sin(0.125)+2(0.125)=0.002182+0.25=0.252182$

When x=0.216 the result is: $y=sin(0.216)+2(0.216)=0.003769+0.432=0.435769$

When x=0.343 the result is: $y=sin(0.343)+2(0.343)=0.005986+0.686=0.691986$

When x=0.512 the result is: $y=sin(0.512)+2(0.512)=0.008936+1.024=1.032936$

When x=0.729 the result is: $y=sin(0.729)+2(0.729)=0.012723+1.458=1.470723$

When x=1 the result is: $y=sin(1)+2(1)=0.017452+2=2.017452$

7 0

3 years ago