Covert 53.4 to a mixed number

lys-0071 [83]

Answer:

7 0

3 years ago

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What is the answer to this

Alborosie

The answer to the question is b

$9 each hour she works + 7.50 fee
$7.50+$9h= $34.50

7 0

3 years ago

You have 42,784 grams of a radioactive kind of curium. If its half-life is 18 years, how much will be left after 72 years?

s344n2d4d5 [400]

Answer:

2,674.14 g

Step-by-step explanation:

Recall that the formula for radioactive decay is

N = N₀ e^(-λt)

where,

N is the amount left at time t

N₀ is the initial amount when t=0, (given as 42,784 g)

λ = coefficient of radioactive decay

= 0.693 ÷ Half Life

= 0.693 ÷ 18

= 0.0385

t = time elapsed (given as 72 years)

e = exponential constant ( approx 2.7183)

If we substitute these into our equation:

N = N₀ e^(-λt)

= (42,787) (2.7183)^[(-0.0385)(72)]

= (42,787) (2.7183)^(-2.7726)

= (42,787) (0.0625)

= 2,674.14 g

6 0

3 years ago

A 140 kg camera is suspended by two wires over a 40 metre wide football field to get shots of the action from above. At one poin

Katen [24]

Answer:

Please red the answer below

Step-by-step explanation:

In order to determine the length of each cable you use the Newton second law for each component of the forces involved in the situation.

For the x component you have:

$T_1cos\theta_1-T_2cos\theta_2=0$ (1)

T1: tension of the first cable = 1500N

T2: tension of the second cable = 800N

θ1: angle between the horizontal and the first cable

θ2: angle between the horizontal and the first cable

For the y component you have:

$T_1sin\theta_1+T_2sin\theta_2-W=0$ (2)

W: weight of the camera = Mg = (140kg)(9.8m/s^2) = 1372N

You can squared both equations (1) and (2) and the sum the two equations:

$T_1^2cos^2\theta_1=T_2^2cos^2\theta_2\\\\T_1^2sin^2\theta_1=T_2^2sin^2\theta_2-2WT_2sin\theta_2+W^2$

Then, you sum the equations:

$T_1^2(cos^2\theta_1+sin^2\theta_1)=T_2^2(sin^2\theta_2+cos^2\theta_2)-2Wsin\theta_2+W^2$ (3)

Next, you use the following identity:

$sin^2\theta+cos^2\theta=1$

and you obtain in the equation (3):

$T_1^2=T_2^2-2WT_2sin\theta_2+W^2\\\\sin\theta_2=\frac{T_2^2-T_1^2+W^2}{2WT_2}=\frac{(800N)^2-(1500)^2+(1372N)^2}{2(800N)(1372N)}=0.066\\\\\theta_2=sin^{-1}0.066=27.23\°$

With this values you can calculate the value of the another angle, by using the equation (1):

$\theta_1=cos^{-1}(\frac{T_2cos(27.23\°)}{T_1})=cos^{-1}(\frac{(800N)(cos27.23\°)}{1500N})\\\\\theta_1=61.69\°$

Now, you can calculate the length of each cable by using the information about the width of the football field. You use the following trigonometric relation:

$l_1cos\theta_1=40-d\\\\l_2cos\theta_2=d\\\\$

d: distance to the right side of the field

By using the cosine law you can fins a system of equation and then you can calculate the values of l1 and l2.

3 0

3 years ago

F(x)=3x+10x and g(x)=2x-4, find (f+g)(x)

djyliett [7]

Answer:

15x - 4

Step-by-step explanation:

Add both of the equations together.

f(x) + g(x)

(3x + 10x) + (2x -4)

Then, combine like terms

3x + 2x + 10x = 15x

15x - 4

8 0

3 years ago

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