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nekit [7.7K]
3 years ago
14

Before a bag of flour can be sold, it must be 40 kilograms but can be within 0.5

Mathematics
1 answer:
Serjik [45]3 years ago
6 0

Answer:

The minimum acceptable weight of the flour bag is 39.5 kg,

and its maximum acceptable weight is 40.5 kg

Step-by-step explanation:

Notice that the selling weight of the bag to be sold (x) needs to differ from 40 kg at most 0.5 kg. Therefore, we can write the following inequality:

|x-40|\leq 0.5

The inequality can be solve once we remove the absolute value symbol, and solve for the unknown "x". Recall that in order to solve it, we need to consider the two possible cases:

a) that the expression within the absolute value symbol is larger than or equal to zero, and b) that the expression is less than zero.

a) If x-40\geq 0 then its absolute value is equal to itself (x-40), and when we remove the absolute value symbols, we get:

x-40\leq 0.5\\x\leq 0.5+40\\x\leq 40.5\,\,kg

which means that the weigh "x" of the flour bag should be smaller or equal than 40.5  kg

b) If x-40, then the absolute value of this is its opposite : -x+40, and the inequality becomes:

-x+40\leq 0.5\\40\leq 0.5+x\\40-0.5\leq x\\39.5\leq x

which means that the weight of the flour bag must be larger than or equal to 39.5 kg

So, the minimum acceptable weight of the flour bag is 39.5 kg, and the maximum acceptable weight is 40.5 kg

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The square ABCD is divided into eight equal parts. The shaded area is 25 cm². What is the area of the square ABCD​
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6 0
3 years ago
A helicopter is flying above a town. the local high school is directly to the east of the helicopter at a 20° angle of depressio
guapka [62]
Answer: 4.5 miles

Explanation:

When you draw the situation you find two triangles.

1) Triangle to the east of the helicopter

a) elevation angle from the high school to the helicopter = depression angle from the helicopter to the high school = 20°

b) hypotensue = distance between the high school and the helicopter

c) opposite-leg to angle 20° = heigth of the helicopter

d) adyacent leg to the angle 20° = horizontal distance between the high school and the helicopter = x

2) triangle to the west of the helicopter

a) elevation angle from elementary school to the helicopter = depression angle from helicopter to the elementary school = 62°

b)  distance between the helicopter and the elementary school = hypotenuse

c) opposite-leg to angle 62° = height of the helicopter

d) adyacent-leg to angle 62° = horizontal distance between the elementary school and the helicopter = 5 - x

3) tangent ratios

a) triangle with the helicpoter and the high school

tan 20° = Height / x ⇒ height = x tan 20°

b) triangle with the helicopter and the elementary school

tan 62° = Height / (5 - x) ⇒ height = (5 - x) tan 62°

c) equal the height from both triangles:

x tan 20° = (5 - x) tan 62°

x tan 20° = 5 tan 62° - x tan 62°

x tan 20° + x tan 62° = 5 tan 62°

x  (tan 20° + tan 62°) = 5 tan 62°

⇒ x = 5 tant 62° / ( tan 20° + tan 62°)

⇒ x = 4,19 miles

=> height = x tan 20° = 4,19 tan 20° = 1,525 miles

4) Calculate the hypotenuse of this triangle:

hipotenuese ² = x² + height ² = (4.19)² + (1.525)² = 19.88 miles²

hipotenuse = 4.46 miles

Rounded to the nearest tenth = 4.5 miles

That is the distance between the helicopter and the high school.
5 0
3 years ago
Read 2 more answers
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