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mario62 [17]
3 years ago
8

15% of $3.50 useing proportion

Mathematics
2 answers:
Otrada [13]3 years ago
3 0

Answer:

0.525

Step-by-step explanation:

\frac{x}{3.50}=\frac{15}{100}

Simora [160]3 years ago
3 0

Answer:

.525

Step-by-step explanation:

Just do 3.50 * 15% and percent means OF 100 so 15/100 is .15

so do 3.50 * .15 to get .525

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The sum of mike and Jordan’s age is 51. The difference in their ages is 17. If mike is older, what are the two men’s ages?
iogann1982 [59]

Answer:

Step-by-step explanation:

You can make equations.

M+J=51

M-J=17

Subtracting first equation from second equation:

2J=34

J=17 Jordan is 17

Therefore, Mike, is 34.

4 0
3 years ago
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
I need help with numbers 5,6,and 8
erastovalidia [21]
5) Obtuse triangle because the top angle is more than 90 degrees
6 0
2 years ago
Subtracting a negative is the same as....
erik [133]
Same as adding together
3 0
2 years ago
Read 2 more answers
Which of the following is equivalent to? 19 ​
11Alexandr11 [23.1K]

Answer:

we dont see the numbers

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
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