Question 1

The measure of RST can be represented by the expression (6x + 12)°.

Valentin [98]

Answer:

The measure of angle RST is $120\°$

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

$m$ -----> equation A

$m$ -----> equation B

equate equation A and equation B

$(6x+12)\°=78\°+(3x-12)\°$

$6x-3x=78\°-24\°$

$3x=54\°$

$x=18\°$

The measure of angle RST is equal to

$m$

8 0

3 years ago

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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.

Ben:

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

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

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

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

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 $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

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:

Whose textbook reaches the ground first and by how many seconds?

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

Express 5/8 as a percent. 1.40% 2.62.5% 3.58% 4.62%

Drupady [299]

Answer:

the answer is #2-62.5%

Step-by-step explanation:

3 0

3 years ago

4,139 divided by 14 equals what?

Anon25 [30]

Answer:

that would be 57,946

3 0

3 years ago

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PLEASE HELP DUE IN 1 HOUR

kobusy [5.1K]

Choosing blue or red: 3 blue + 4 red = 7 marbles
3 blue + 2 green + 4 red + 1 yellow = 10 marbles

Probability blue or red marble: 7/10

Choosing a marble that isn't blue or red: 2 green + 1 yellow = 3 marbles
Total marbles: 3 blue + 2 green + 4 red + 1 yellow = 10 marbles

Probability non-blue or red marble: 3/10

5 0

3 years ago