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3 years ago

36 workers can complete a piece of work in 12 days how many workers should be removed to complete the same work in 24 days

Mathematics

2 answers:

meriva3 years ago

3 0

His answer is going to have to be 19393939 because i like men

Anit [1.1K]3 years ago

3 0

18. They multiplied the amount of days by 2, so you divide the amount of workers by 2. This is reverse/opposite operations. I hope this helps :)

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Write an equation for: The quotient of a number and eight is five less than the number.

strojnjashka [21]

$\frac{n}{8} = n - 5$

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3 years ago

Simplify the number into simplest radical form. Use the factor tree to help determine the factors. StartRoot 96 Endroot StartRoo

adell [148]

Answer:

$[tex]4608\sqrt{3}$ [/tex]

Step-by-step explanation:

1. $\sqrt{96} *\sqrt{6}* 2*\sqrt{6}*4 *\sqrt{6} *4*\sqrt{3}$

2. $\sqrt{2^{5} *3} \sqrt{6} *2\sqrt{6}*4\sqrt{6}*4\sqrt{3}$ factoring 96

since $\sqrt{2^{5}*3 } = \sqrt{2^{5} } \sqrt{3}$

3. $\sqrt{2^{5} } \sqrt{3}\sqrt{6} *2\sqrt{6}*4\sqrt{6}*4\sqrt{3}$

using exponent rule - $(a^{b}) ^{c} = a^{bc}$

$\sqrt{2^{5} } = 2^{5/2}$

4. $2^{5/2}\sqrt{3}\sqrt{2*3} *2\sqrt{6}*4\sqrt{6}*4\sqrt{3}$

doing some simple simplification and $4=2^{2}$ and 6=2*3

5. $2^{5/2} \sqrt{3} \sqrt{2} \sqrt{3} *2\sqrt{2} \sqrt{3} *2^{2}\sqrt{2} \sqrt{3}*4\sqrt{3}$

collecting the roots on one side and applying exponent rule

6. $\sqrt{3} \sqrt{3}\sqrt{3} \sqrt{3}\sqrt{2} \sqrt{2}\sqrt{2} *2^{5/2+1+2+2} \sqrt{3}$

Applying exponents rule on all $\sqrt{3}$ and $\sqrt{2}$

7. $2^{1/2+1/2+1/2} *2^{5/2+1+2+2}*3^{1/2+1/2+1/2+1/2+1/2}$

combining all powers of 2

8. $2^{1/2+1/2+1/2+5/2+1+2+2}*3^{1/2+1/2+1/2+1/2+1/2}$

Simplifying

9. $2^{9} *3^{2}\sqrt{3}$

10. $512*9\sqrt{3}$

11. $4608\sqrt{3}$

8 0

3 years ago

Read 2 more answers

Find the point P on the graph of the function y=√x closest to the point (9,0)

Sphinxa [80]

Answer:

$\displaystyle \frac{17}{2}$ .

Step-by-step explanation:

Let the $x$ -coordinate of $P$ be $t$ . For $P\!$ to be on the graph of the function $y = \sqrt{x}$ , the $y$ -coordinate of $\! P$ would need to be $\sqrt{t}$ . Therefore, the coordinate of $P \!$ would be $\left(t,\, \sqrt{t}\right)$ .

The Euclidean Distance between $\left(t,\, \sqrt{t}\right)$ and $(9,\, 0)$ is:

$\begin{aligned} & d\left(\left(t,\, \sqrt{t}\right),\, (9,\, 0)\right) \\ &= \sqrt{(t - 9)^2 +\left(\sqrt{t}\right)^{2}} \\ &= \sqrt{t^2 - 18\, t + 81 + t} \\ &= \sqrt{t^2 - 17 \, t + 81}\end{aligned}$ .

The goal is to find the a $t$ that minimizes this distance. However, $\sqrt{t^2 - 17 \, t + 81}$ is non-negative for all real $t\!$ . Hence, the $\! t$ that minimizes the square of this expression, $\left(t^2 - 17 \, t + 81\right)$ , would also minimize $\sqrt{t^2 - 17 \, t + 81}\!$ .

Differentiate $\left(t^2 - 17 \, t + 81\right)$ with respect to $t$ :

$\displaystyle \frac{d}{dt}\left[t^2 - 17 \, t + 81\right] = 2\, t - 17$ .

$\displaystyle \frac{d^{2}}{dt^{2}}\left[t^2 - 17 \, t + 81\right] = 2$ .

Set the first derivative, $(2\, t - 17)$ , to $0$ and solve for $t$ :

$2\, t - 17 = 0$ .

$\displaystyle t = \frac{17}{2}$ .

Notice that the second derivative is greater than $0$ for this $t$ . Hence, $\displaystyle t = \frac{17}{2}$ would indeed minimize $\left(t^2 - 17 \, t + 81\right)$ . This $t\!$ value would also minimize $\sqrt{t^2 - 17 \, t + 81}\!$ , the distance between $P$ $\left(t,\, \sqrt{t}\right)$ and $(9,\, 0)$ .

Therefore, the point $P$ would be closest to $(9,\, 0)$ when the $x$ -coordinate of $P\!$ is $\displaystyle \frac{17}{2}$ .

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3 years ago

Multiply two-and-two-thirds by one-fourth.

a_sh-v [17]

The answer is two thirds (or 2/3)

Hope this helps! :)

6 0

4 years ago

How much will joseph have to pay

koban [17]

40% of 740 is 444, and 30% of 444 is 133.2 ,so the answer is $133.2

6 0

2 years ago

Read 2 more answers