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Gala2k [10]
3 years ago
11

Evaluate the following expression when a = 5 and b = 6 3a +10b

Mathematics
1 answer:
Anni [7]3 years ago
6 0
The answer is 75 I believe :)
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PLEASE HELP
lubasha [3.4K]

Answer:

(a+b)^2+(a-b)^2=2(a^2+b^2)

expanding:

a² + 2ab + b² + a² -2ab + b² = 2a² + 2b²

2a² + 2b² = 2a² + 2b²

7 0
3 years ago
Read 2 more answers
Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be
Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of t=0.6482

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

h(t)=-16t^2 +v_{o}t+s       Eqn(1).

Where:

h(t) : is the height range as a function of time

v_{o}   : is the initial velocity

s     : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

h(t)=-16t^2 + v_{o}t + 40        Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

<u>Ben:</u>

We know that v_{o}=60ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+60t+40        Eqn.(3)

<u>Josh:</u>

We know that v_{o}=48ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+48t+40       Eqn. (4).

<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em>h(t)=0<em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

Thus we have for Ben, Eqn. (3) gives:

h(t)=0-16t^2+60t+40=0

Using the quadratic expression to find the roots of the quadratic we have:

t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec

Since time can only be positive we reject the t_{2} solution and we keep that Ben's book took t=4.3276 seconds to reach the ground.

Similarly solving for Josh we obtain

t_{1}=3.6794sec\\t_{2}=-0.6794sec

Thus again we reject the negative and keep the positive solution, so Josh's book took t=3.6794 seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482

So to answer the original question:

<em>Whose textbook reaches the ground first and by how many seconds?</em>

  • Josh's textbook reached the ground first
  • Josh's textbook reached the ground first by a difference of t=0.6482

3 0
3 years ago
Any help would be appreciated! <br> 7/9+ 8/9 as a mixed number
katovenus [111]

Answer:

15/9 :))

is your answerrr

7 0
3 years ago
Read 2 more answers
Which is equal to cos47°?
Murrr4er [49]

Which of these is equal to cos 47°?

a. cos 43°

b. sin 43°

c. tan 43°

d. cos 133°

B.....

cos (47°)= sin (90°-47°)=sin (43°) [complementary angles]

Answer B

7 0
2 years ago
Hello please help i’ll give brainliest
MAXImum [283]

Answer:

Just a little puppy

hope this helped

Step-by-step explanation:

8 0
2 years ago
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