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3 years ago

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devon records 4 hours of reality shows on her dvr she records comedy shows 3/8 of that amount of time. Devon watches all the rea

lity and comedy shows in half hour sittings. Find the number of hours of comedy shows that Devon records

2 answers:

rodikova [14]3 years ago

8 0

Answer:

$1\frac{1}{2}$ hours

Step-by-step explanation:

Devon records reality shows on her dvr for the time of = 4 hours

She records comedy shows $\frac{3}{8}$ of the amount of time.

The time of recording comedy shows = $\frac{3}{8}$ of 4 hours

= $\frac{3}{8}$ × 4

= $\frac{12}{8}$

= $\frac{3}{2}$

= $1\frac{1}{2}$ hours

The number of hours of comedy shows that Devon records is $1\frac{1}{2}$ hours or 1.5 hours.

scZoUnD [109]3 years ago

6 0

Devon records 3/8 of 4 hours for comedy shows, which is 90 mins worth (1 hour and a half).

Please vote my answer brainliest. thanks!

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Answer:

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Step-by-step explanation:

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3 years ago

Jose has $45 to spend at the mall. He buys a toy for $18.63, two candy bars at $0.47 EACH, and a shirt for $12.64 (All prices in

GREYUIT [131]

Answer:

$12.79

Step-by-step explanation:

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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

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irina1246 [14]

Answer:

1. ASA

2. AAS

3. AAS

4. SSS

5. SSS

6. not sure which triangles are number 6

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2 years ago

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kow [346]

Answer:

-3/10

Step-by-step explanation:

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3 years ago