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katovenus [111]
4 years ago
6

Which graph is the solution to lxl > 10?

Mathematics
1 answer:
ale4655 [162]4 years ago
5 0

\huge\text{Hey there!}

\huge\text{Note}\rightarrow \leq \& \geq \ = \huge\text{closed circle}\\ < \& > = \huge\text{(o)(p)(e)(n)(e)(d) circle}

\huge\text{So, we know that A or C could be our answer}.

\huge\text{Eliminate option B and D}

→ \huge\text{Find the absolute value of the problem}

→ \huge\text{You can now know that  it could be:}  

\huge\text{x}\large\huge{\ >}\huge\text{ 10 or x }< \huge\text{ -10}

→ \huge\text{Possible answer \#1: x }>\huge\text{10}\\\\\huge\text{Possible answer \#2: x}

→ \boxed{\boxed{\huge\text{Answer: C}}}  

\huge\text{Option A. would be too small} \huge\checkmark

\text{Good luck on your assignment and enjoy your day!}

~\frak{LoveYourselfFirst:)}

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Answer:

I'm guessing y is numerator and 5 is denominator. so y^2/25

Step-by-step explanation:

so if you have the fraction y/5 and you need to square it or raise to the second power you get (y/5)^2. now using the properties of exponents when you get a fraction and need to raise it to a power you square the denominator and numerator which gets you y^2/25. now let's check if it works we can use 5 for y so (5/5)^2 is equal to 25/25 = 1. or (10/5)^2 which is 2^2 = 4. or 100/25 which is 4. so the equation works for the expression.

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Please help to solve this.<br> Find sin5x,if sinx+cosx=1,4
MrMuchimi

Answer:

Hello,

Step-by-step explanation:

we must first remember that:

\boxed{sin(5x)=5sin(x)-20*sin^3(x)+16*sin^5(x)}\\\\

Let say t=sin(x)

sin(x)+cos(x)=1.4\\\\t+\sqrt{1-t^2}=1.4\\\sqrt{1-t^2}=1.4-t\ we\ square\\1-t^2=1.4^2+t^2-2.8*t\\2t^2-2.8*t+0.96=0\\\Delta=2.8^2-4*2*0.96=0.16=0.4^2\\\\t=\dfrac{2.8-0.4}{4} =0.6\ or\ t=\dfrac{2.8+0.4}{4}=0.8\\\\t^3=0,216\ or\ t^3=0,512\\\\t^5=0,07776\ or\ t^5=0,32768\\\\\boxed{sin(5x)=5*0.6-20*0.216+16*0.07776=-0,07584}\\\\or\\\\\boxed{sin(5x)=5*0.8-20*0.512+16*0.32768=-0,99712}

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Use the graph of the function y = 4x to answer the following questions.
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Step-by-step explanation:

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