Asked by Brady
I will appreciate it if anyone can help me with the following:
1. The optimal height h of the letters of a message printed on pavement is given by the formula . Here, h= (0.00252d^(9/4))/e d is the distance of the driver from the letters and e is the height of the driver’s eye above the pavement. All of the distances are in meters. Explain how to find h when d = 90 m and e = 1.4 m.
2.Simplify.
(27a^(-6) )^(-2/3)
For 2 I got 1/(9a^4 )as my answer, but can someone again please check this.
For 1 I think the answer is:
First, you replace the variables with 90 and 1.4 to get this equation: (0.00252∙〖90〗^(9/4))/1.4. Next, you turn the 〖90〗^(9/4)into ∜(〖90〗^9 ). Then, I turn ∜(〖90〗^9 )into 3∜(〖10〗^9 ), and simplify 3∜(〖10〗^9 )to 9∜10. Next, I divide o.oo252 by 1.4 to get 0.0018. Then, my next equation is: 0.0018∙9∜10. Finally, I multiply 0.0018 and 9 together to get 0.0162, and my final answer is 0.0162∜10.
If anyone can answer or help me for these two problems, please do. I will appreciate it even more if you can answer and/or check the problems as soon as possible. I have been waiting and trying to answer it for a long time. My mom wants me to get my lesson done on this topic today, and I am stuck at this point, which is really close to the end. I am willing to go step by step for each one until I get the right answer. Also, thank you if anyone can help or answer these two problems.
1. The optimal height h of the letters of a message printed on pavement is given by the formula . Here, h= (0.00252d^(9/4))/e d is the distance of the driver from the letters and e is the height of the driver’s eye above the pavement. All of the distances are in meters. Explain how to find h when d = 90 m and e = 1.4 m.
2.Simplify.
(27a^(-6) )^(-2/3)
For 2 I got 1/(9a^4 )as my answer, but can someone again please check this.
For 1 I think the answer is:
First, you replace the variables with 90 and 1.4 to get this equation: (0.00252∙〖90〗^(9/4))/1.4. Next, you turn the 〖90〗^(9/4)into ∜(〖90〗^9 ). Then, I turn ∜(〖90〗^9 )into 3∜(〖10〗^9 ), and simplify 3∜(〖10〗^9 )to 9∜10. Next, I divide o.oo252 by 1.4 to get 0.0018. Then, my next equation is: 0.0018∙9∜10. Finally, I multiply 0.0018 and 9 together to get 0.0162, and my final answer is 0.0162∜10.
If anyone can answer or help me for these two problems, please do. I will appreciate it even more if you can answer and/or check the problems as soon as possible. I have been waiting and trying to answer it for a long time. My mom wants me to get my lesson done on this topic today, and I am stuck at this point, which is really close to the end. I am willing to go step by step for each one until I get the right answer. Also, thank you if anyone can help or answer these two problems.
Answers
Answered by
Ms. Sue
Perhaps you need a personal tutoring session with your instructor.
Answered by
Steve
#1 (0.00252d^(9/4))/e
= (0.00252*90^(9/4))/1.4
90^(1/4) = ∜90 = 3.0801
90^(9/4) = 3.0801^9 = 24948.5693
So, we now have
0.00252 * 24948.5693 / 1.4 = 44.9074
If you visit wolframalpha.com you can play around with arbitrarily complex expressions, just to see how things combine.
You went wrong starting here:
Then, I turn ∜(〖90〗^9 )into 3∜(〖10〗^9 )
Where did that 3 come from?
∜(〖90〗^9 ) = ∜(90^8 * 90) = 90^(8/4) * ∜90 = 90^2 * ∜90 = 8100∜90
You seem to lose track of the pieces when factoring stuff out.
#2
(27a^(-6) )^(-2/3)
27^(-2/3) * a^(-6 * -2/3)
1/9 a^4
a^4 / 9
= (0.00252*90^(9/4))/1.4
90^(1/4) = ∜90 = 3.0801
90^(9/4) = 3.0801^9 = 24948.5693
So, we now have
0.00252 * 24948.5693 / 1.4 = 44.9074
If you visit wolframalpha.com you can play around with arbitrarily complex expressions, just to see how things combine.
You went wrong starting here:
Then, I turn ∜(〖90〗^9 )into 3∜(〖10〗^9 )
Where did that 3 come from?
∜(〖90〗^9 ) = ∜(90^8 * 90) = 90^(8/4) * ∜90 = 90^2 * ∜90 = 8100∜90
You seem to lose track of the pieces when factoring stuff out.
#2
(27a^(-6) )^(-2/3)
27^(-2/3) * a^(-6 * -2/3)
1/9 a^4
a^4 / 9
Answered by
Brady
Thank you Steve. I will try to do better during these type of problems, I will even do it your way. Thank you again!
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