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3 years ago

If a = 5 and b=2 then find the value of - 2ab and - 3a+b

Mathematics

2 answers:

attashe74 [19]3 years ago

7 0

Answer:

-2ab = -20
-3a + b = -13

Concept:

Here, we need to understand the idea of evaluation.

When encountering questions that gave you an expression with variables, then stated: "If x = a, y = b, z = c" (a, b, c are all constants), this means you should substitute the value given for each variable back to the expression.

Solve:

<h2>PART I: Find the value of -2ab</h2>

Given information

a = 5

b = 2

Given expression

-2ab

Substitute values into the expression

-2 (5) (2)

Simplify by multiplication

=-10 (2)

= $\boxed{-20}$

--------------------------------------------------------

<h2>PART II: Find the value of -3a + b</h2>

Given information

a = 5

b = 2

Given expression

-3a + b

Substitute values into the expression

-3 (5) + (2)

SImplify by multiplication

=-15 + (2)

Simplify by addition

= $\boxed{-13}$

Hope this helps!! :)

Please let me know if you have any questions

dusya [7]3 years ago

3 0

-2ab = -2(5)(2) = -20
-3a+b = -3(5)+2 = -13

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3 years ago

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3 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:

Camera 2nd has to cover the maximum angle, i.e. $78.70^\circ$ .

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

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

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

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

Camera 2nd has to cover the maximum angle, i.e. $78.70^\circ$ .

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4 years ago