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monitta
3 years ago
14

HELP ASAP!!

Mathematics
2 answers:
blondinia [14]3 years ago
7 0

Answer:

84 square inches

Step-by-step explanation:

First, separate this shape into two shapes you can easily find the area of. You can do this by turning it into a rectangle and a rhombus.

Say the rectangle has measurements of 5 by 6 inches. The area of that rectangle will be 30 square inches. Save this value to add on later.

The rhombus now has a height of 4 (because the side length 10 minus the side length 6 equals 4), and two bases by the values of 12 and 15. The area of a rhombus can be found with the equation:

a = 1/2(b1 + b2) x h

Simply plug in those values to find the area of the rhombus, 54. Now add the area of the rhombus and the area of the rectangle to get 84 square inches as your final area.

Monica [59]3 years ago
3 0
B because it makes sense
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How much will it cost in Canadian dollars to purchase €2000 at a bank that charges a 2.2% commission on the transaction? Use an
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$2,874.25 Canadian dollar

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The first thing to do here is to calculate the amount of the commission.

That would be 2.2% of €2,000

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Now the total cost to pay = selling price + amount in commissions = €2,000 + €44 = €2,044

But we need our answer in the Canadian dollars

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4 0
3 years ago
How do you solve 5(r+9)−2(1−r)=1<br> need quickly
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5 0
2 years ago
Three security cameras were mounted at the corners of a triangles parking lot. Camera 1 was 110 ft from camera 2, which was 137
Nata [24]

Answer:

<em>Camera 2nd has to cover the maximum angle, i.e. </em>78.70^\circ.

Step-by-step explanation:

Please have a look at the triangular park represented as a triangle \triangle ABC with sides

a = 110 ft

b = 158 ft

c = 137 ft

1st camera is located at point C, 2nd camera at point B and 3rd camera at point A respectively.

We can use law of cosines here, to find out the angles \angle A, \angle B, \angle C

As per Law of cosine:

cos C = \dfrac{a^{2}+b^2-c^2 }{2ab}\\cos B = \dfrac{a^{2}+c^2-b^2 }{2ac}\\cos A = \dfrac{b^{2}+c^2-a^2 }{2bc}

Putting the values of a,b and c to find out angles \angle A, \angle B, \angle C.

cos C = \dfrac{110^{2}+158^2-137^2 }{2\times 110 \times 158}\\\Rightarrow cos C = \dfrac{12100+24964-18769 }{24760}\\\Rightarrow cos C =0.526\\\Rightarrow C = 58.24^\circ

cos B = \dfrac{110^{2}+137^2-158^2 }{2\times 110 \times 137}\\\Rightarrow cos B = \dfrac{12100+18769 -24964}{30140}\\\Rightarrow cos B = \dfrac{5905}{30140}\\\Rightarrow cos B =0.196\\\Rightarrow B = 78.70^\circ

cos A = \dfrac{158^{2}+137^2-110^2 }{2\times 158 \times 137}\\\Rightarrow cos A = \dfrac{24964+18769-12100}{43292}\\\Rightarrow cos A = \dfrac{31633}{43292}\\\Rightarrow cos A = 0.731\\\Rightarrow A = 43.05^\circ

<em>Camera 2nd has to cover the maximum angle</em>, i.e. 78.70^\circ.

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