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bezimeni [28]
3 years ago
5

What is the shaded area of this shape?

Mathematics
1 answer:
OLEGan [10]3 years ago
3 0

Answer:

A=150

Step-by-step explanation:

you can split the shape into a square and triangle. the square's area is 15 x 6 =  90. the triangles area is (1/2)    (14-6) (15)= 60. then you add them together and get 150

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A set of furniture was sold for GH cedis 3000 at a profit of 25%. Find the cost price.
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<h2>2400GH Cedis </h2>

Step-by-step explanation:

Selling \:price = 3000gh Cedis\\Profit \% = 25\%\\Cost\:price =?\\\\CP =\frac{ ( Selling\:price \times 100 )}{ ( 100 + percentage profit)}  \\\\C.P = \frac{3000\times 100 }{100+25}\\\\ C.P = \frac{300000}{125} \\\\Cost\: Price = 2400gh Cedis

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A surveyor leaves her base camp and drives 42km on a bearing of 032degree she then drives 28km on a bearing of 154degree,how far
ValentinkaMS [17]

Answer:

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

Step-by-step explanation:

The final position of the surveyor is represented by the following vectorial sum:

\vec r = \vec r_{1} + \vec r_{2} + \vec r_{3} (1)

And this formula is expanded by definition of vectors in rectangular and polar form:

(x,y) = r_{1}\cdot (\cos \theta_{1}, \sin \theta_{1}) + r_{2}\cdot (\cos \theta_{2}, \sin \theta_{2}) (1b)

Where:

x, y - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.

r_{1}, r_{2} - Length of each vector, in kilometers.

\theta_{1}, \theta_{2} - Bearing of each vector in standard position, in sexagesimal degrees.

If we know that r_{1} = 42\,km, r_{2} = 28\,km, \theta_{1} = 32^{\circ} and \theta_{2} = 154^{\circ}, then the resulting coordinates of the final position of the surveyor is:

(x,y) = (42\,km)\cdot (\cos 32^{\circ}, \sin 32^{\circ}) + (28\,km)\cdot (\cos 154^{\circ}, \sin 154^{\circ})

(x,y) = (35.618, 22.257) + (-25.166, 12.274)\,[km]

(x,y) = (10.452, 34.531)\,[km]

According to this, the resulting vector is locating in the first quadrant. The bearing of the vector is determined by the following definition:

\theta = \tan^{-1} \frac{10.452\,km}{34.531\,km}

\theta \approx 16.840^{\circ}

And the distance from the camp is calculated by the Pythagorean Theorem:

r = \sqrt{(10.452\,km)^{2}+(34.531\,km)^{2}}

r = 36.078\,km

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

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