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Dvinal [7]
3 years ago
12

PLEASE HELP DUE AT 11:59 For the inequality, find two values for r that make the inequality true: -5x<2

Mathematics
2 answers:
Gnesinka [82]3 years ago
4 0

Answer:

b,e

Step-by-step explanation:

-5x -2= 10 false

-5x3=-15 true

-5x-5=25 false

-5x-6=30 false

-5x5=-25 true

jonny [76]3 years ago
3 0

\large\textbf{Hey there!}

\mathbf{-5x < 2}\\\\\huge\textbf{DIVIDE -5 to BOTH SIDES}\\\mathbf{\dfrac{-5}{-5} < \dfrac{2}{-5}}\\\huge\textbf{CANCEL out: }\huge\boxed{\mathbf{\dfrac{-5}{-5}}}\huge\textbf{ because it give you 1}\\\huge\textbf{KEEP: }\huge\boxed{\mathbf{\dfrac{2}{-5}}}\huge\textbf{ because it help what is being}\\\huge\textbf{compared to the x-value}\\\large\textbf{NEW EQUATION: x }\mathbf{ > \dfrac{2}{-5}}\\\\\huge\textbf{SIMPLIFY IT!}

\huge\textbf{Thus, your answer is: \boxed{\mathsf{x > - \dfrac{2}{5}}}}\huge\checkmark

\large\textbf{Good luck on your assignment \& enjoy your day!}

<h3>~\frak{Amphitrite1040:)}</h3>
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iragen [17]

The height of the rocket above the ground after t seconds is given by the equation :  H= -16t^2+Vt+h , where V is the initial velocity and h is the initial height.

Given that, V= 500 ft/second and h= 20 ft

So, the equation will become:  H= -16t^2 +500t+20

A) For finding the height of the rocket 3 seconds after the launch, we will <u>plug t=3 into the above equation</u>. So....

H= -16(3)^2+500(3)+20\\ \\ H= -16(9)+1500+20\\ \\ H= -144+1500+20=1376

So, the height of the rocket 3 seconds after the launch is 1376 feet.

B) When the rocket at a height of 400 feet, then <u>we will plug H= 400</u>

400=-16t^2+500t+20\\ \\ 16t^2-500t-20+400=0\\ \\ 16t^2-500t+380=0\\ \\ 4(4t^2-125t+95)=0\\ \\ 4t^2-125t+95=0

Using quadratic formula, we will get......

t= \frac{-(-125)+/-\sqrt{(-125)^2-4(4)(95)}}{2(4)}\\ \\ t= \frac{125+/-\sqrt{14105}}{8}\\ \\ t= 30.4705... \\ and \\ t= 0.7794...

So, after 0.7794...seconds and 30.4705...seconds the rocket is at a height of 400 feet above the ground.

C) The time duration that the rocket remains in the air means we need to find <u>the time taken by the rocket to reach the ground</u>. When it reaches the ground, then H=0. So.....

0=-16t^2+500t+20\\ \\ -4(4t^2-125t-5)=0\\ \\ 4t^2-125t-5=0

Using <u>quadratic formula</u>, we will get.....

t= \frac{-(-125)+/-\sqrt{(-125)^2-4(4)(-5)}}{2(4)}\\ \\ t= \frac{125+/-\sqrt{15705}}{8}\\ \\ t=31.2899...\\ and\\ t= -0.0399...

<em>(Negative value is ignored as time can't be in negative)</em>

So, the rocket will remain in the air for 31.2899... seconds.

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Answer:

In the attachment!!

<em>Hope this helps!</em>

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