All categories

3 years ago

C(3) = 4/3x + -4 solve plsss, i appreciate it

Mathematics

1 answer:

pogonyaev3 years ago

8 0

The equation given is:

c(x) = 4/3x + (-4)

Plug in 3 and solve:

c(3) = 4/3(3) + (-4)

c(3) = 12/3 + (-4)

c(3) = 4 + (-4)

Answer: c(3) = 0

Hope I helped!

-Char

You might be interested in

There are 15 students in the class and 8 are boys. What is the ratio of girls to students in the room?

cricket20 [7]

Answer:

7/15

Step-by-step explanation:

hope this helps?:)

8 0

3 years ago

Which of the following functions are homomorphisms?

Vikentia [17]

Part A:

Given $f:Z \rightarrow Z,$ defined by $f(x)=-x$

$f(x+y)=-(x+y)=-x-y \\ \\ f(x)+f(y)=-x+(-y)=-x-y$

but

$f(xy)=-xy \\ \\ f(x)\cdot f(y)=-x\cdot-y=xy$

Since, f(xy) ≠ f(x)f(y)

Therefore, the function is not a homomorphism.

Part B:

Given $f:Z_2 \rightarrow Z_2,$ defined by $f(x)=-x$

Note that in $Z_2$ , -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular $f(x)=x$

$f(x+y)=x+y \\ \\ f(x)+f(y)=x+y$

and

$f(xy)=xy \\ \\ f(x)\cdot f(y)=xy$

Therefore, the function is a homomorphism.

Part C:

Given $g:Q\rightarrow Q$ , defined by $g(x)= \frac{1}{x^2+1}$

$g(x+y)= \frac{1}{(x+y)^2+1} = \frac{1}{x^2+2xy+y^2+1} \\ \\ g(x)+g(y)= \frac{1}{x^2+1} + \frac{1}{y^2+1} = \frac{y^2+1+x^2+1}{(x^2+1)(y^2+1)} = \frac{x^2+y^2+2}{x^2y^2+x^2+y^2+1}$

Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.

Part D:

Given $h:R\rightarrow M(R)$ , defined by $h(a)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)$

$h(a+b)= \left(\begin{array}{cc}-(a+b)&0\\a+b&0\end{array}\right)= \left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right) \\ \\ h(a)+h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)+ \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)=\left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right)$

but

$h(ab)= \left(\begin{array}{cc}-ab&0\\ab&0\end{array}\right) \\ \\ h(a)\cdot h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)\cdot \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)= \left(\begin{array}{cc}ab&0\\-ab&0\end{array}\right)$

Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.

Part E:

Given $f:Z_{12}\rightarrow Z_4$ , defined by $\left([x_{12}]\right)=[x_4]$ , where $[u_n]$ denotes the lass of the integer $u$ in $Z_n$ .

Then, for any $[a_{12}],[b_{12}]\in Z_{12}$ , we have

$f\left([a_{12}]+[b_{12}]\right)=f\left([a+b]_{12}\right) \\ \\ =[a+b]_4=[a]_4+[b]_4=f\left([a]_{12}\right)+f\left([b]_{12}\right)$

and

$f\left([a_{12}][b_{12}]\right)=f\left([ab]_{12}\right) \\ \\ =[ab]_4=[a]_4[b]_4=f\left([a]_{12}\right)f\left([b]_{12}\right)$

Therefore, the function is a homomorphism.

7 0

3 years ago

A football team has a squad of 24 players,

lora16 [44]

Answer: 1/3

Explanation:

1/12 of the squad are suspended:

1/12 x 24 = 24/12 = 2

So 2 player are suspended

24 - 2 = 22 player

And some are injured leaving 14 left to play:

22 - 14 = 8 players injured

So 8/24 is the fraction of injured player.

But 8/24 = 1/3 so we can say 1/3 of the whole squad are injured.

4 0

3 years ago

Read 2 more answers

In a study of the impact of smoking on birth weight, researchers analyze birth weights (in grams) for babies born to 189 women w

kompoz [17]

Answer:

c. Smoking is associated with lower birth weights. When smokers are compared to nonsmokers, we are 95% confident that the mean weight of babies born to nonsmokers will be between 76.5 grams to 486.9 grams more than the mean birth weight of babies born to smokers.

Step-by-step explanation:

The difference in mean birth weights (nonsmokers minus smokers) is 281.7 grams with a margin of error of 205.2 grams with 95% confidence.

Then, we know that the 95% confidence interval is (76.5, 486.9)

$LL=M-MOE=281.7-205.2=76.5\\\\UL=M-MOE=281.7+205.2=486.9$

a. We are 95% confident that smoking causes lower birth weights by an average of between 76.5 grams to 486.9 grams.

False, it interprets the confidence interval as a probability for individual cases. The confidence interval is a range for the population mean.

b. There is a 95% chance that if a woman smokes during pregnancy her baby will weigh between 76.5 grams to 486.9 grams less than if she did not smoke.

False, it interprets the confidence interval as a probability for individual cases. The confidence interval is a range for the population mean.

c. Smoking is associated with lower birth weights. When smokers are compared to nonsmokers, we are 95% confident that the mean weight of babies born to nonsmokers will be between 76.5 grams to 486.9 grams more than the mean birth weight of babies born to smokers.

True.

d. With such a large margin of error, this study does not suggest that there is a difference in mean birth weights when we compare smokers to nonsmokers.

There is enough evidence, as the lower bound of the confidence is positive. This means that there is only a probability of 0.05/2=0.025 that the true mean weight difference is smaller than 76.5 grams.

6 0

3 years ago

PLEASE HELP

antiseptic1488 [7]

Answer:

2.Ellipse

Step-by-step explanation:

The given equation represents an ellipse.

Standard form of ellipse is given as:

$\frac{x^2}{a^2} +\frac{y^2}{b^2} = 1\\\\\implies \frac{x^2}{16} - \frac{y^2}{4} = 1\\$

3 0

3 years ago