+0  
 
0
75
3
avatar+59 

Please help me with these questions, I need urgent help FAST!

 

1) Let line l\(l_1\) be the graph of 5x + 8y = -9. Line L2 is perpendicular to line L1 and passes through the point (10, 10). If line L2 is the graph of the equation y=mx +b, then find m+b .

 

2) Find all real numbers x  such that \(​​​​3x - 7 \le 5x +9\). Give your answer as an interval.

 

3) Find all real numbers s such that \(-2(s - 7) > 4s + 8\) . Give your answer as an interval.

 

4) Without using a calculator, order the following numbers from least to greatest: \(\begin{align*} A &= \frac{2^{1/2}}{4^{1/6}}\\ B &= \sqrt[12]{128}\vphantom{dfrac{2}{2}}\\ C&= \left( \frac{1}{8^{1/5}} \right)^2\\ D&= \sqrt{\frac{4^{-1}}{2^{-1} \cdot 8^{-1}}}\\ E&= \sqrt[3]{2^{1/2} \cdot 4^{-1/4}}.\vphantom{dfrac{2}{2}}\end{align*}\)
Give your answer as a list of capital letters separated by commas. For example, if you think that E < B < C< D< A , then you would answer E,B,C,D,A.

 

Sorry for so many questions, help as fast as you can, please! Thanks to anyone who answers! I'll definitely give you extra likes if you answer!

Web2.0CalcUser123  Nov 17, 2018

Best Answer 

 #1
avatar+316 
+4

1)\(5x + 8y = -9 <=> y=-\frac{5}{8}x-\frac{9}{8} \) 

λ1=-\(\frac{5 }{8}\) 

L2 is perpendicular to line L1 so λ1*λ2=-1 <=> -5/8*λ2=-1

λ2=\(\frac{8}{5}\)

so the l2: y-y1=λ2(x-x1)  x1=10,y1=10 because  passes through the point (10, 10)

so \(y-10 = \frac{8}{5}(x-10)<=> y=\frac{8}{5}x-\frac{80}{5}+10<=> y=\frac{8}{5}x-\frac{30}{5}\)

m=8/5,b=-30/5

m+b=\(\frac{8}{5}-\frac{30}{5}=\frac{-22}{5}\)

 

2)\(3x-7≤ 5x+9<=> 3x-5x≤9+7 <=> -2x≤16 <=> x≥ 8\)

\(x≥ 8\)

 

3)\(-2s+14>4s +8<=> 14-8 > 4s+2s <=> 6>6s <=> 1>s <=> s<1\)

\(s<1\)

 

4) https://ibb.co/ffXqkL

 

Hope this helps! 

Dimitristhym  Nov 18, 2018
edited by Dimitristhym  Nov 18, 2018
edited by Dimitristhym  Nov 18, 2018
 #1
avatar+316 
+4
Best Answer

1)\(5x + 8y = -9 <=> y=-\frac{5}{8}x-\frac{9}{8} \) 

λ1=-\(\frac{5 }{8}\) 

L2 is perpendicular to line L1 so λ1*λ2=-1 <=> -5/8*λ2=-1

λ2=\(\frac{8}{5}\)

so the l2: y-y1=λ2(x-x1)  x1=10,y1=10 because  passes through the point (10, 10)

so \(y-10 = \frac{8}{5}(x-10)<=> y=\frac{8}{5}x-\frac{80}{5}+10<=> y=\frac{8}{5}x-\frac{30}{5}\)

m=8/5,b=-30/5

m+b=\(\frac{8}{5}-\frac{30}{5}=\frac{-22}{5}\)

 

2)\(3x-7≤ 5x+9<=> 3x-5x≤9+7 <=> -2x≤16 <=> x≥ 8\)

\(x≥ 8\)

 

3)\(-2s+14>4s +8<=> 14-8 > 4s+2s <=> 6>6s <=> 1>s <=> s<1\)

\(s<1\)

 

4) https://ibb.co/ffXqkL

 

Hope this helps! 

Dimitristhym  Nov 18, 2018
edited by Dimitristhym  Nov 18, 2018
edited by Dimitristhym  Nov 18, 2018
 #2
avatar+656 
0

If you want to use "-->" or "==>" in LaTex, it is $\Rightarrow$. 

PartialMathematician  Nov 19, 2018
 #3
avatar+59 
0

Big shout out to Dimitristhym for answering my questions, but how do you write the answer to number 2) in regular form? I can't get the x_> 8 things right. Thank you all, though!

Web2.0CalcUser123  Nov 20, 2018

2 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.