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Licemer1 [7]
3 years ago
11

If s represents speed write an inequality to represent how fast a person can legally dive a vehicle in a school zone with a spee

d limit of 25 mph ?
Mathematics
1 answer:
ELEN [110]3 years ago
5 0
The phrase “inequality” tells you that you will be using inequality signs, greater than, less than, greater than or equal to, or less than or equal to. Those signs look like this (not ordered) ≤ ≥ < >
So now we have to consider the speed limit 25mph, and remember s is used to represent how fast you are going. We can start be setting up 25_s and now we must determine what sign to use. Does it make sense to go faster than the speed limit? No it doesn’t, it’s a limit. what about equal to, does it make sense to go at the speed limit? Yes you can travel 25mph in a 25mph zone. So your speed will be less than or equal to 25, because you know you can’t go faster than that but you can travel at that number. So it looks like:
25 ≥ s.
If you need help understanding why I used that particular inequality sign message me for help.
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Please help me with the picture attached I am so confused :(
sleet_krkn [62]

Answer:

ez

Step-by-step explanation:

14/15 - 9/15 = 1/3

1/3 = 5/15

3 0
3 years ago
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4. Find the value of each variable for the right triangles. Make sure all answers are in reduced radical form. Show your work.
lara [203]
Use Pythagorean's, in figure 1: t^2+12^2=13^2. t=\sqrt(13^2-12^2)=\sqrt(25)=5.
In figure 2: 6^2+9^2=x^2. x=\sqrt(36+81)=\sqrt(117)=3*\sqrt(13).
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3 years ago
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PLEASE HELPPPPPPPPPPP
Marizza181 [45]
The answer would be C
8 0
3 years ago
What is the sum?
gavmur [86]

Answer:

The answer is option D.

Step-by-step explanation:

First we must first find the LCM

the LCM of x² - 9 and x + 3 is x² - 9

So we have

\frac{3}{ {x}^{2} - 9 }  +  \frac{5}{x + 3}  =   \frac{3 + 5(x - 3)}{ {x}^{2}  - 9}  \\  \\  =  \frac{3 + 5x - 15}{(x + 3)(x - 3)}    \\  \\  = \frac{5x - 12}{(x +  3)(x - 3)}

Hope this helps you

5 0
3 years ago
How can I factorise (2x cubed - 5x + 3) ?
PilotLPTM [1.2K]
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therefore\\x=\frac{-2\pm\sqrt{2^2-4(2)(-3)}}{2(2)}=\frac{-2\pm\sqrt{4+24}}{4}=\frac{-2\pm\sqrt{28}}{4}=\frac{-2\pm\sqrt{4\cdot7}}{4}=\frac{-2\pm2\sqrt7}{4}\\\\=\frac{-1\pm\sqrt7}{2}\\\\so,\ the\ answer:\\\\(x-1)\cdot2\left(x-\frac{-1-\sqrt7}{2}\right)\left(x-\frac{-1+\sqrt7}{2}\right)\\\\=\boxed{(x-1)(2x+1+\sqrt7)\left(x+\frac{1-\sqrt7}{2}\right)}=\boxed{\frac{1}{2}(x-1)(2x+1+\sqrt7)(2x+1-\sqrt7)}
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3 years ago
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