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katovenus [111]

3 years ago

14

On average people spend 1/4 of the time they sleep in a dream state. If Maxwell slept 10 hours last night, how much time did he

spend dreaming? Write your answer in simplest form.

2 answers:

DedPeter [7]3 years ago

7 0

Hey there, Matthew!

So 1/4 turns to 0.25

Now multiply 0.25 x 10 = 2.5

So Maxwell spent 2.5 hours dreaming.

Hope I was able to help!

saveliy_v [14]3 years ago

5 0

Maxwell dreamed a total of 2.5 hours, here's how I got my answer.

10 divided by 1/4 = 2.5

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Show that (A − B) − C is not necessarily equal to A − (B − C).

Leni [432]

(3-2)-1=0

3-(2-1)=2

Boom! Done

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

I need help with my algebra 2. I also want to know how to solve this.

Lisa [10]

Add 1 to both sides:

$\sqrt{x+3} = x+1$

In cases like this, we have to remember that a root is always positive, so we can square both sides only assuming that

$x+1 \geq 0 \iff x \geq -1$

Under this assumption, we square both sides and we have

$x+3 = (x+1)^2 \iff x+3 = x^2+2x+1 \iff x^2+x-2 = 0$

The solutions to this equation are

$x = -2,\ x=1$

But since we can only accept solutions greater than -1, we discard $x=-2$ and accept $x=1$ .

In fact, we have

$x=-2 \implies \sqrt{-2+3}-1=0\neq -2$

and

$x=1 \implies \sqrt{4}-1=1$

which is the only solution.

4 0

3 years ago

A circular garden has a circumference of 113 yards. Lars is digging a straight line along a diameter of the garden at a rate of

natima [27]

Answer:3hrs 59min

Step-by-step explanation:

πd=perimeter

(113÷3.14)

=35.99

9yards=1hr

35.99yards=?

35.99÷9=3hrs 59min

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3 years ago

What term refers slope

Serga [27]

In y = mx+b, m is the term that refers to slope

hope this helps

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3 years ago