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slavikrds [6]
3 years ago
14

SAR.

Mathematics
1 answer:
ozzi3 years ago
5 0

Answer:

its an sectet of jet ski cant tell sorry

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A motorcyclist leaves City A and rides at a constant speed towards City B, which is 225 km away. After 1.5 hours, the motorcycli
rewona [7]

Answer:

50 km/hr

Step-by-step explanation:

OK, this is how I solved this problem.

Let r = original rate of speed

     t = time after the stop to finish the trip

Time for trip without stopping is 225/r

(1)     1.5 + .5 + t = 225/r      This is a time equation

(2)    1.5r + t(r + 10) = 225     This is a distance equation

(1)       2 + t = 225/r            (2)      1.5r + tr + 10t = 225

         2r + tr = 225

                tr = 225 - 2r                 1.5r + 225 - 2r + 10(225/r - 2) = 225

                t = 225/r - 2

                                                   -.5r + 225 + 2250/r - 20 = 225

                                                     -5r + 2250 + 22500/r - 200 = 2250

                                                     -5r^2 + 2250r + 22500 - 200r = 2250r

                                                       5r^2 + 200r - 22500 = 0

                                                       5(r^ + 40r - 4500) = 0

                                                           5(r - 50)(r + 90) = 0

                                                                 r = 50   or    r = -90

The rate cannot be negative, so the original speed 50 km/hr

This not an easy problem and I hope you were able to follow my work

5 0
3 years ago
Read 2 more answers
-63<br> —— = -7 what’s the error <br> -9
STatiana [176]

Answer:

One of the negatives must go

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Hi, can someone please help, and possibly explain the answer please.
mr_godi [17]

Answer:

x=\frac{-(-2)\±\sqrt{(-2)^2-4(3)(0)} }{2(3)}

Step-by-step explanation:

Quadratic formula: x=\frac{-b\±\sqrt{b^2-4ac} }{2a} when the equation is 0=ax^2+bx+c

The given equation is 1=-2x+3x^2+1. Let's first arrange this so its format looks like y=ax^2+bx+c:

1=-2x+3x^2+1

1=3x^2-2x+1

Subtract 1 from both sides of the equation

1-1=3x^2-2x+1-1\\0=3x^2-2x+0

Now, we can easily identify 3 as a, -2 as b and 0 as c. Plug these into the quadratic formula:

x=\frac{-b\±\sqrt{b^2-4ac} }{2a}\\x=\frac{-(-2)\±\sqrt{(-2)^2-4(3)(0)} }{2(3)}

I hope this helps!  

8 0
2 years ago
I WILL GIVE BRAINLIEST TO CORRECT ANSWER
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180 students have a pet
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Please help me with number 9
Mrrafil [7]
The answer is 8 pints
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