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meriva
1 year ago
6

In one week, we sold 184 baskets at $3.25 each. The following week, we reduced the price of the basket to $2.

Mathematics
1 answer:
8_murik_8 [283]1 year ago
7 0

The number of baskets that were sold for $2 is 299 baskets.

<h3>What is a word problem?</h3>

A word problem is a method we used in denoting mathematical expressions and variables and it can be solved with the use of fractions, ratios, algebra, and arithmetic operations as the case may be.

From the given information:

  • If 184 baskets is sold for = $3.25;
  • and (x) baskets is sold for = $2

The number of baskets sold in the second week is obtained as follows

i.e.

x = (184 × 3.25)/2

x = 299 basket were sold for $2  

Learn more about solving word problems here:

brainly.com/question/13818690

#SPJ1

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1. A buoy floats 19 yards from the eastern most point of a boat and 15 yards from the western most point of a second boat. The a
Black_prince [1.1K]

The laws of cosines and law of sines can be used given that two sides

and an included angle, or two angles a side are known.

Response:

1. The other angles in the triangle formed by the buoy are approximately;

  • <u>31.1° and 40.9°</u>

2. Distance of the helicopter from the first island is approximately;

  • <u>14.5 miles</u>

<h3>How is the Law of Sines and Cosines used?</h3>

Given parameters are;

Distance of the buoy from the easternmost point of a boat = 19 yards

Distance of the buoy from the westernmost point of the other boat = 15 yards

Angle formed from the buoy to the two boats = 108°

Distance between the two boats, <em>d</em>, is given by the law of cosines, as follows;

d² = 19² + 15² - 2 × 19 × 15 × cos(108°) = 586 - 570·cos(108°)

d = √(586 - 570·cos(108°))

By the law of Sines, we have;

\dfrac{d}{sin(108^{\circ})} = \mathbf{\dfrac{15}{sin(Angle \ formed \ from \ the \ boat \ on \ the \ West, \ \theta_1)}}

Which gives;

sin(\theta_1) = \mathbf{ \dfrac{15 \times sin(108^{\circ})}{\sqrt{586 - 570 \cdot cos(108^{\circ})} }}

The o

\theta_1 = arcsin \left( \dfrac{15 \times sin(108^{\circ})}{\sqrt{586 - 570 \cdot cos(108^{\circ})} } \right) \approx   \mathbf{31.1^{\circ}}

The other angles formed in the triangle containing the buoy are;

  • θ₁ ≈ <u>31.1</u>
  • θ₂ ≈ 180° - 108° - 31.1° ≈<u> 40.9°</u>

2. Distance between the two islands = 20 miles

Angle of elevation with one island = 15°

Angle of elevation with the second island = 35°

Required:

The mileage (distance travelled) of the helicopter.

Solution:

Let <em>A</em> represent the island that has an angle of elevation to the helicopter

of 15°, and let <em>B</em> represent the other island.

Angle formed by the helicopter and the two island, θ, is found as follows;

θ = 180° - (15° + 35°) = 130°

By the Law of Sines, we have;

\dfrac{20}{sin(130^{\circ})} = \mathbf{ \dfrac{Distance \ from \  island \ A }{sin(35^{\circ})}}

Which gives;

Distance \ of \ helicopter \ from \  island \ A = \mathbf{ \dfrac{20}{sin(130^{\circ})} \times sin(35^{\circ})}

Mileage \ from \ island \ A =  \dfrac{20}{sin(130^{\circ})} \times sin(35^{\circ}) \times cos(15^{\circ}) \approx 14.5

  • The mileage of the helicopter from the first island is approximately <u>14.5 miles</u>

Learn more about the Law of Sines and Cosines here:

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Step-by-step explanation:

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Answer:

Hello,

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0.75 is the decimal format.


5 0
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None of those is.  The expression   4 In x + In 3 – In r  

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