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2 years ago

Use the diagram to answer the questions

Mathematics

1 answer:

IRISSAK [1]2 years ago

8 0

Answer:

1. B. BD

2. C. 15

3. A. 2

Step-by-step explanation:

1. From the figure,

∠DAB = ∠DAC

Therefore, BD is the bisector of ∠ABC.

2. From the figure, IG = GH

3x = 5x - 10

10 = 5x - 3x

2x = 10

x = 5

So, GH = 5x - 10

= 5(5) - 10

= 25 - 10

= 15

Therefore, GH = 15.

3. From the figure,

∠DAB = ∠DAC

10y = 8y + 4

10y - 8y = 4

2y = 4

y = 2

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Answer:

Step-by-step explanation:

The 4 is in the hundredths place, so it represents 4 hundredths, or 4 x 1/100.

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2 years ago

Old McDonald had a farm. In one pen the old farmer had chickens and bulls together. In that pen there were 40 animals in total,

SVETLANKA909090 [29]

Answer:

There are 15 numbers of chicken and 25 numbers of bulls.

Step-by-step explanation:

The pen had chickens and bulls. The pen contains 40 animals in total. The number of legs in the pen is 130 .

Let

a = number of chicken

b = number of bull

a + b = 40................(i)

A chicken has 2 legs and a bull has 4 legs.

The total legs can be represented as

2a + 4b = 130...........(ii)

combine the equation

a + b = 40................(i)

2a + 4b = 130...........(ii)

a = 40 - b

insert the value of a in equation(ii)

2(40 - b) + 4b = 130

80 - 2b + 4b = 130

80 + 2b = 130

2b = 130 - 80

2b = 50

divide both sides by 2

b = 50/2

b = 25

Insert the value of b in equation (i)

a + b = 40

a + 25 = 40

a = 40 - 25

a = 15

There are 15 numbers of chicken and 25 numbers of bulls.

8 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 $\angle A, \angle B, \angle C$ .

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

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

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