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2 years ago

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Carletta and Johnathan are entrepreneurs making mother boards for computers that are being manufactured by their friend, Dewey S

asser. Carletta can make one mother board in 12 hours while it takes Johnathan 16 hours to make one mother board. Dewey suggested they could produce mother boards faster if they combined forces and worked together to build each mother board. What will be the time, in hours, that Carletta and Johnathan can build a mother board when working together? (Round your answer to the nearest hour.)

1 answer:

nikitadnepr [17]2 years ago

3 0

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Someone help me please!

andriy [413]

Answer:

To solve the first inequality, you need to subtract 6 from both sides of the inequality, to obtain 4n≤12. This can then be cancelled down to n≤3 by dividing both sides by 4. To solve the second inequality, we first need to eliminate the fraction by multiplying both sides of the inequality by the denominator, obtaining 5n>n^2+4. Since this inequality involves a quadratic expression, we need to convert it into the form of an^2+bn+c<0 before attempting to solve it. In this case, we subtract 5n from both sides of the inequality to obtain n^2-5n+4<0. The next step is to factorise this inequality. To factorise we must find two numbers that can be added to obtain -5 and that can be multiplied to obtain 4. Quick mental mathematics will tell you that these two numbers are -4 and -1 (for inequalities that are more difficult to factorise mentally, you can just use the quadratic equation that can be found in your data booklet) so we can write the inequality as (n-4)(n-1)<0. For inequalities where the co-efficient of n^2 is positive and the the inequality is <0, the range of n must be between the two values of n whereby the factorised expresion equals zero, which are n=1 and n=4. Therefore, the solution is 1<n<4 and we can check this by substituting in n=3, which satisfies the inequality since (3-4)(3-1)=-2<0. Since n is an integer, the expressions n≤3 and n<4 are the same. Therefore, we can write the final answer as either 1<n<4, or n>1 and n≤3.

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2 years ago

I need this problem solved

Ahat [919]

Answer:

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Step-by-step explanation:

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1 year ago

Is anybody else here to help me ??

Akimi4 [234]

Answer:

$\cot(x)+\cot(\frac{\pi}{2}-x)$

$\cot(x)+\tan(x)$

$\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}$

$\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}]$

$\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}]$

$\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}]$

$\csc(x)[\frac{1}{\cos(x)}]$

$\csc(x)[\sec(x)]$

$\csc(x)[\csc(\frac{\pi}{2}-x)]$

$\csc(x)\csc(\frac{\pi}{2}-x)$

Step-by-step explanation:

I'm going to use $x$ instead of $\theta$ because it is less characters for me to type.

I'm going to start with the left hand side and see if I can turn it into the right hand side.

$\cot(x)+\cot(\frac{\pi}{2}-x)$

I'm going to use a cofunction identity for the 2nd term.

This is the identity: $\tan(x)=\cot(\frac{\pi}{2}-x)$ I'm going to use there.

$\cot(x)+\tan(x)$

I'm going to rewrite this in terms of $\sin(x)$ and $\cos(x)$ because I prefer to work in those terms. My objective here is to some how write this sum as a product.

I'm going to first use these quotient identities: $\frac{\cos(x)}{\sin(x)}=\cot(x)$ and $\frac{\sin(x)}{\cos(x)}=\tan(x)$

So we have:

$\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}$

I'm going to factor out $\frac{1}{\sin(x)}$ because if I do that I will have the $\csc(x)$ factor I see on the right by the reciprocal identity:

$\csc(x)=\frac{1}{\sin(x)}$

$\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

Now I need to somehow show right right factor of this is equal to the right factor of the right hand side.

That is, I need to show $\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)}$ is equal to $\csc(\frac{\pi}{2}-x)$ .

So since I want one term I'm going to write as a single fraction first:

$\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)}$

Find a common denominator which is $\cos(x)$ :

$\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}$

$\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}$

$\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}$

By the Pythagorean Identity $\cos^2(x)+\sin^2(x)=1$ I can rewrite the top as 1:

$\frac{1}{\cos(x)}$

By the quotient identity $\sec(x)=\frac{1}{\cos(x)}$ , I can rewrite this as:

$\sec(x)$

By the cofunction identity $\sec(x)=\csc(x)=(\frac{\pi}{2}-x)$ , we have the second factor of the right hand side:

$\csc(\frac{\pi}{2}-x)$

Let's just do it all together without all the words now:

$\cot(x)+\cot(\frac{\pi}{2}-x)$

$\cot(x)+\tan(x)$

$\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}$

$\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})$

$\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}]$

$\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}]$

$\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}]$

$\csc(x)[\frac{1}{\cos(x)}]$

$\csc(x)[\sec(x)]$

$\csc(x)[\csc(\frac{\pi}{2}-x)]$

$\csc(x)\csc(\frac{\pi}{2}-x)$

7 0

3 years ago

Look at this triangle.<br> 8 cm<br> 22 cm<br> Work out length AB.

Dmitry_Shevchenko [17]

Would the answer be 176? I’m not sure but i’m curious

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2 years ago

Read 2 more answers

If you were to fill a graduated cylinder to 150mL, then drop a metal object in that raises the water level to 250mL, what is the

fiasKO [112]

Answer:

100 mL

Step-by-step explanation:

Subtract 150 from 250 to get the answer, hope this helped!

4 0

2 years ago