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3 years ago

What is 28; 50% increase

Mathematics

1 answer:

saveliy_v [14]3 years ago

6 0

42, because 50% or 1/2 of 28 is 14. Increasing it would be addinng it to the original number so 28+12=42

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Georgia needs to buy some cheese. An 8-ounce package of

const2013 [10]

Answer:

neither buy is better, both buys are $2.28571428/ounce

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3 years ago

What is the sum of the first 47 terms of the sequence? -215, -201, -187, -173, ...

Andru [333]

Answer:

5029

Step-by-step explanation:

There is a common difference between consecutive terms, that is

d = - 201 - (- 215) = - 187 - (- 201) = - 173 - (- 187) = 14

This indicates the sequence is arithmetic with sum to n term

= [ 2a₁ + (n - 1)d ]

where a₁ is the first term and d the common difference

Here a₁ = - 215 and d = 14 , then

= [ (2 × - 215) + (46 × 14) ]

= 23.5 (- 430 + 644)

= 23.5 × 214

= 5029

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2 years ago

The aquarium has 3 more yellow fish than green fish. 60 percent of the fish are yellow. How many green fish are in the aquarium?

kompoz [17]

9514 1404 393

Answer:

Step-by-step explanation:

If 60% are yellow, the other 40% are green. Then the difference between yellow and green is 60% -40% = 20% of the total number of fish. That difference also happens to be half of the number of green fish:

20% = (1/2)(40%)

So, the number of green fish is 2×3 = 6.

_____

If you wanted to, you could write a couple of equations:

y - g = 3

y = 0.60(y +g)

Using the first equation to write an expression for y, we get ...

y = g +3

Using this in the second equation, we have ...

g +3 = 0.6(g +3 +g)

g +3 = 1.2g +1.8 . . . . eliminate parentheses

1.2 = 0.2g . . . . . . . . subtract g+1.8

1.2/0.2 = g = 6 . . . . . divide by 0.2

There are 6 green fish in the aquarium.

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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.

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3 years ago

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3 years ago