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2 years ago

12

La distancia de Urano al Sol es 2 870 990 000 km y la distancia de la Tierra al Sol es 1,496 × 108 km. Aproximadamente, ¿cuántas

veces está contenida la distancia de la Tierra al Sol en la distancia de Urano al Sol?

1 answer:

Leni [432]2 years ago

6 0

Answer:

2,7204x 10^9

Step-by-step explanati

2870000000-

149600000

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PLEASE HELP ASAP: A particle is moving with velocity v(t) = t2 – 9t + 18 with distance, s measured in meters, left or right of z

dimaraw [331]

1) $\frac{v(8)-v(0)}{8 - 0} = \frac{10-18}{8} = -1\frac{m}{s}$ .
2) $v(5) = 5^2-9*5+18 = 25-45+18 = -2\frac{m}{s}$ .
3) The particle is moving right when the velocity function is positive: $0\ \textless \ t\ \textless \ 3$ or $6\ \textless \ t\ \textless \ 8$ .
4) When $0\ \textless \ t\ \textless \ \frac{9}{2}$ the particle is slowing down because the acceleration is close to zero $\Rightarrow$ the particle is speeding up when acceleration is increasing away from zero: $\frac{9}{2}\ \textless \ t\ \textless \ 8$ .
5) $(\frac{1}{8})* \int\limits^8_0 {t^2-9t+18dt}=\frac{1}{8}*(\frac{t^3}{3}-(\frac{9}{2}*t^2+18t)_{0}^{8}= \\=(\frac{1}{8})*(\frac{8^3}{3}-(\frac{9}{2})*8^2+18*8)=\frac{8^2}{3}-(\frac{9}{2})*8+18=3\frac{1}{3} \frac{m}{s}$ .

3 0

2 years ago

Your cell phone plan charges a base charge each month plus a charge per daytime minute of usage and December used 510 daytime mi

algol13

Answer:

In February, 423 daytime minutes is used

Step-by-step explanation:

Let the base plan charges be x

And cost per daytime minute be y

In December,

x + 510y = 92.25------------------(1)

In January,

x + 397y = 77.56---------------------(2)

Subtracting eq(2) from eq(1)

x + 510y = 92.25

x + 397y = 77.56

-------------------------------

0 + 113y = 14.69

-------------------------------

y = \frac{14.69}{113}

y = 0.13----------------------------------(3)

Substituting (3) in (1)

x + 510(0.13) = 92.25

x + 66.3 = 92.25

x = 92.25 - 66.3

x = 25.95

So In February

base plan + (daytime minute)(cost per daytime minute) = 80.9

25.95 + (daytime minute)(0.13) = 80.9

(daytime minute)(0.13) = 80.9 - 25.95

(daytime minute)(0.13) = 54.95

(daytime minute) = $\frac{54.95}{0.13}$

daytime minutes = 422.69

daytime minute $\approx 423$

3 0

2 years ago

hichkok12 [17]

Answer: 345.02

Step-by-step explanation:

8 0

1 year ago

Find two complex numbers that have a sum of i10 a different of -4and a product of -29

lianna [129]

<h2>-2+5i and 2+5i</h2>

Step-by-step explanation:

Let the complex numbers be $a+ib\textrm{ and }c+id$ .

Given, sum is $10i$ , difference is $-4$ and product is $-29$ .

$(a+c)+i(b+d)=10i$ ⇒ $a+c=0,b+d=10$

$(a-c)+i(b-d)=-4$ ⇒ $a-c=-4,b-d=0$

$a=-2,c=2,b=5,d=5$

$(a+ib)(c+id)=(-2+5i)(2+5i)=-4-25=-29$

Hence, all three equations are consistent yielding the complex numbers $-2+5i\textrm{ and }2+5i$ .

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1 year ago

The track team is trying to reduce their time for a relay race. First they reduce their time by 2.1 minutes. Then they are able

konstantin123 [22]

Answer: 15.7 minutes

Step-by-step explanation:

Let x be the time in the beginning (in minutes).

Given: The track team is trying to reduce their time for a relay race.

First they reduce their time by $t_1=$ 2.1 minutes.

Then they are able to reduce that time by $t_2=$ 10

If their final time is 3.96 minutes, then

$x-t_1-t_2=3.6\\\Rightarrow\ x=3.6+t_1+t_2\\\Rightarrow\ x=3.6+2.1+10\\\Rightarrow\ x= 15.7$

Hence, their beginning time was 15.7 minutes.

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2 years ago