En el juego de domino son 28 fichas con la cara hacia abajo cual es la probabilidad de sacar la ficha 6-7

7^-5 times 7^3 equals what, I need help asap please

djverab [1.8K]

Your answer is 7^(- 2) or $\frac{1}{ 7^{2} }$ or $\frac{1}{49}$

5 0

2 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.

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

Un terreno rectangular cuyo ancho mide 20 mts menos que su largo. Su área del terreno es de 150 metros cuadrados. ¿Determina las

IrinaVladis [17]

Answer:

you are a good friend Sayed im not sure if I can get it done before I get my license back and the whole time we go back and be out in a bit I guess that was the best thing I ever heard of the beast's I have

7 0

2 years ago

The first term of an arithmetic sequence is 14. The 25th term is 206. What is the common difference of the arithmetic sequence?

larisa [96]

an = a1+ d(n-1)

a1 =14

an=206 when n=25

206 = 14 + d (25-1)

206 = 14 + d * 24

subtract 14 from each side

192 = 24d

divide by 24 on each side

d=8

The common difference is 8

3 0

2 years ago

Read 2 more answers

Please help me I appreciate it

miss Akunina [59]

Answer:

m∠SRT = 33° ∵ m∠QRT = m∠SRT

Step-by-step explanation:

Given the diagram,

RT bisects ∠SRQ, and m∠QRT = 33°

RT bisecting ∠SRQ means RT will bisect ∠SRQ into two equal angles, which are:

m∠QRT = m∠SRT

as

m∠QRT = 33°

so

m∠SRT = 33° ∵ m∠QRT = m∠SRT

6 0

2 years ago