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brilliants [131]
3 years ago
5

Identify the GCF of 12a^4b^3 + 8a^3b^2. 4a^3b^2 2a^3b 4a^2b 8a^3b^2

Mathematics
1 answer:
Bess [88]3 years ago
7 0
A is the answer you seek
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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
Mr. SMITH had 4 daughters, each daughter had a brother ... How many sons does Mr. Smith have?
Zielflug [23.3K]

He tells us that the 4 daughters have a brother, so, since they belong to the same family, the sisters are the same because they have only one brother.

Answering our question:

or Mr. Smith has 4 daughters and also 1 son.

6 0
4 years ago
Read 2 more answers
Help someone plz I need it
kicyunya [14]

Answer:

1) There are double number of pennies than there are quarters

2) For every 2 pennies, there is one quarter

3) There are 2 less quarters compared to the pennies

Hope dis helped!

Step-by-step explanation:

plz mark as brianliest if correct!!!

3 0
3 years ago
Read 2 more answers
A light bulb consumes 3600 watts- hours per day. How many watt - hours does it consume in 5 days and 6 hours?
Serjik [45]

Answer:

13500/3600= 3 days and 3/4 of a day

5 0
3 years ago
Read 2 more answers
(3rd time Asking) Please help me with these questions!!!
poizon [28]

(a) Water balloon 1 has height at time t

h_1(t)=-16t^2-20t+110

and water balloon 2 has height

h_2(t)=-16t^2+75

Both water balloons are moving straight downward, so their distance is equal to

|h_1(t)-h_2(t)|=|(-16t^2-20t+110)-(-16t^2+65)|=|35-20t|

(b) When both water balloons are at the same height, we have h_1(t)=h_2(t). So you could set both quadratic expressions equal to each other and solve for t. Or you can use the fact that, when they're at the same height, the distance between them is 0. So using the expression from part (a), we have

|35-20t|=0\implies 35-20t=0\implies 35=20t\implies t=\dfrac{35}{20}

or 1.75 seconds after being thrown/dropped.

At this time, they are both at a height of h_1(1.75)=h_2(1.75):

h_2(1.75)=-16(2.25)^2+75=26

feet.

(c) The second and fourth statements are true.

4 0
4 years ago
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