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kari74 [83]
4 years ago
14

If a = 3 and b = 5.what is the value of a2 + B2?

Mathematics
2 answers:
mariarad [96]4 years ago
3 0

Answer:

16

Step-by-step explanation:

If you replace the variables with what they equal and multiply then you should get 6+10 which equals 16.

Marysya12 [62]4 years ago
3 0

Answer:

The answer I received for your equation is 16


Step-by-step explanation:

Step 1) Replace your variables with the given number and rewrite your problem:  3 * 2 + 5 * 2 = __

Step 2) Solve from left to right: 3 * 2 = 6 , 5 * 2 = 10

Step 3) Add your two answers: 6 + 10 = 16

Your final answer should be 16.


Hope I helped, Good luck in your future studies :)


-A


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<h3>What is an angle?</h3>

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Which of these expressions is equivalent to (12/5) ?
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Rachel runs 2km to her bus stop, and then rides 4.5 km to school. On average, the bus is 45 km/h faster than Rachel's average ru
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Answer:6.042km/h 6km/h approximately

Step-by-step explanation:

First off we have to know the formula relating speed, distance and time which is

Speed = distance/time

Now we are looking for Rachel's running speed

We are to find Rachel's running speed, so let's label is x

We are given that the distance Rachel runs to her bus stop is 2km

We were not given the time she uses to run to the bus stop

So let's label the time Rachel uses to run to her bus stop as y

So from the formula speed = distance/time

We have x = 2/y

Now we are told that the speed the bus uses to get to school is 45km/h faster than her speed used to run

So speed of bus = 45 + x

And the overall time for the whole journey is 25mins, changing this to hours, because the speed details given is in km/h we divide 25 by 60 which will give 0.417

Now if the total time is 0.417 hours, and we labeled the time for Rachel to run to the bus as y, so the time for the time for the bus to get to school will be 0.417 - y

We are also told the bus rides for 4.5km to school

So adding this together to relate the speed, distance and time of the bus with the formula speed = distance/time

We get 45 + x = 4.5/(0.417 - y)

So we have two equations

x = 2/y (1)

45+x = 4.5/(0.417-y) (2)

So putting (1) in (2) we have

45 + (2/y) = 4.5/(0.417-y)

Expanding further

(45y + 2)/y = 4.5(0.417-y)

Cross multiplying

(45y + 2)(0.417 - y) = 4.5y

Opening the brackets

18.765y - 45y2 + 0.834 - 2y = 4.5y

Collecting like terms

-45y2 + 18.765y -2y - 4.5y + 0.834 = 0

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I'm using almighty formula her

For solving

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x = (-b +-root(b2-4ac)/2a

For our own equation, we are finding y

From our our quadratic equation

a = 1, b=-0.273, c = -0.834

you = (-(-0.273)+-root(-0.273-4(1)(-0.019))/2(1)

y = (273+-root(0.151))/2

y = (0.273+0.389)/2 or (0.273-0.389)/2

y = 0.331 or -0.085

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Then we put y = 0.331 in (1)

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x = 2/0.331

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x = 6km/h approximately

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