The ideal mass for a piece of chocolate is 2.5 ounces. The actual mass for a production can vary by, at most, 0.08

Mathematics

1 answer:

Nady [450]2 years ago

8 0

Answer:

$2.42\,oz \leq m_{real} \leq 2.58\,oz$

Step-by-step explanation:

The inequality must be the absolute value of the difference between real mass and ideal mass less or equal to 0.08 ounces. That is to say:

$|m_{real} - m_{ideal}| \leq 0.08\,oz$

$|m_{real} - 2.5\,oz| \leq 0.08\,oz$

There are two extreme cases:

$m_{real,max} - 2.5\,oz = 0.08\,oz$ and $m_{real, min} - 2.5\,oz = -0.08\,oz$

$m_{real, max} = 2.58\,oz$ and $m_{real,min} = 2.42\,oz$

Range of chocolate masses must be:

$m_{real, min} \leq m_{real} \leq m_{real,max}$

$2.42\,oz \leq m_{real} \leq 2.58\,oz$

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A ladder 10 m long,leans against a vertical wall at an angle of 70° to the ground.if the ladder slips down the wall 4m,find,corr

lina2011 [118]

Answer:

(a) the new angle the ladder makes with the ground is $32.7^o$

(b) the ladder slipped back about 5 meters

Step-by-step explanation:

Notice that the ladder doesn't change its length in the process.

So let's start from the initial situation , finding the distance from the ground at which the ladder touches the wall when the angle with the ground is 70^o. Notice that this situation is represented by a right angle triangle with the right angle between the wall and the ground (see attached image), and that we can use the sine function to find the side opposite to the 70 degree angle:

$sin(70^o)=\frac{opposite}{hypotenuse} \\sin(70^o)=\frac{h}{10}\\h=10\, sin(70^o) \approx 9.4 \,\,m$

therefore 9.4 meters is approximately the height at which the ladder touches the wall initially.

Now, if the tip of the ladder goes down the wall 4 meters, it is now at 9.4 m - 4 m = 5.4 m from the ground. We can therefore use again the sine function to solve for the new angle:

$sin(x)=\frac{opposite}{hypotenuse} \\sin(x)=\frac{5.4}{10} \\sin(x)=0.54\\x=arcsin(0.54)\\x= 32.7^o$

To answer the second question we need to find the original distance from the wall that the bottom of the ladder was originally, and for that we can use the cosine function:

$cos(70^o)=\frac{adjacent}{hypotenuse} \\cos(70^o)=\frac{x}{10}\\x=10\,cos(70^o)\\x\approx 3.4 \,\,m$