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mario62 [17]
3 years ago
15

Someone please help I got a movie on the line

Mathematics
1 answer:
andrew-mc [135]3 years ago
8 0

Answer:

(-6, -3)?

Step-by-step explanation:

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Match each transformation to the best description of the transformation.
zhannawk [14.2K]

Answer:

c a b

Step-by-step explanation:

brailiest pls

5 0
3 years ago
Help me
Mandarinka [93]

Answer:

$3!! hope this helps!!!

3 0
3 years ago
Crystal left her running shoes at school yesterday. Today she walked 44 miles to school to get her shoes, she ran home along the
Darina [25.2K]

Answer:

We can conclude that her walking speed is 2.1 miles per hour.

Step-by-step explanation:

We have the relation:

Speed = distance/time.

Here we know:

She walked for 44 miles.

And she ran along the same route, so she ran for 44 miles.

The total time of travel is 22 hours, so if she ran for a time T, and she walked for a time T', we must have:

T + T' = 22 hours.

If we define: S = speed runing

                     S' = speed walking

Then we know that:

"and she ran 33 miles per hour faster than she walked."

Then:

S = S' + 33mi/h

Then we have four equations:

S'*T' = 44 mi

S*T = 44 mi

S = S' + 33mi/h

T + T' = 22 h

We want to find the value of S', the speed walking.

To solve this we should start by isolating one of the variables in one of the equations.

We can see that S is already isoalted in the third equation, so we can replace that in the other equations where we have the variable S, so now we will get:

S'*T' = 44mi

(S' + 33mi/H)*T = 44mi

T + T' = 22h

Now let's isolate another variable in one of the equations, for example we can isolate T in the third equation to get:

T = 22h - T'

if we replace that in the other equations we get:

S'*T' = 44mi

(S' + 33mi/h)*(  22h - T') = 44 mi

Now we can isolate T' in the first equation to get:

T' = 44mi/S'

And replace that in the other equation so we get:

(S' + 33mi/h)*(  22h -44mi/S' ) = 44 mi

Now we can solve this for S'

22h*S' + (33mi/h)*22h + S'*(-44mi/S')  + 33mi/h*(-44mi/S') = 44mi

22h*S' + 726mi - 44mi - (1,452 mi^2/h)/S' = 44mi

If we multiply both sides by S' we get:

22h*S'^2 + (726mi - 44mi)*S' - (1,425 mi^2/h) = 44mi*S'

We can simplify this to get:

22h*S'^2 + (726mi - 44mi - 44mi)*S' - (1,425 mi^2/h) = 0

22h*S'^2 + (628mi)*S' - ( 1,425 mi^2/h) = 0

This is just a quadratic equation, the solutions for S' are given by the Bhaskara's equation:

S' = \frac{-628mi  \pm \sqrt{(628mi)^2 - 4*(22h)*(1,425 mi^2/h)}  }{2*22h} \\S' = \frac{-628mi \pm 721 mi }{44h}

Then the two solutions are:

S' = (-628mi - 721mi)/44h = -30.66 mi/h

But this is a negative speed, so this has no real meaning, and we can discard this solution.

The other solution is:

S' = (-628mi + 721mi)/44h = 2.1 mi/h

We can conclude that her walking speed is 2.1 miles per hour.

7 0
3 years ago
Graph this please any math expert please thank you
Pani-rosa [81]

Answer:

we cant see the picture but if i could i would have solved it

Step-by-step explanation:

we cant see the picture n

3 0
3 years ago
my school bag contains 9 books each mass 2/9 kg and 5 folder each mass 7/25 kg what is the total mass of books and folders in my
ololo11 [35]
9*2/9= 18/9= 2kg
5*7/25= 35/25= 1 10/25= 1 2/5 kg
1 2/5kg+2kg= 3 2/5kg or 3.4kg
total mass equals 3 and 2/5kg or 3.4kg
5 0
3 years ago
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