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2 years ago

Pls answer my question from the above picture

Mathematics

2 answers:

steposvetlana [31]2 years ago

4 0

Step-by-step explanation:

I HOPE THIS THIS HELPS U.

liq [111]2 years ago

4 0

Answer:

Step-by-step explanation:

$a^{m}* a^{n}=a^{m+n}\\\\\\(\frac{-1}{2})^{-19}*(\frac{-1}{2})^{8}=(\frac{-1}{2})^{-2x +1}\\\\(\frac{-1}{2})^{-19+8}=(\frac{-1}{2})^{-2x+1}\\\\\\(\frac{-1}{2})^{-11}=(\frac{-1}{2})^{-2x+1}$

As bases are equal in boht sides, we can compare the exponents.

-2x + 1 = -11

Subtract 1 form both sides

-2x = -11 - 1

-2x = -12

Divide both sides by -2

x = -12/-2

x = 6

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15 POINTS! What is the correct answer for Dean's problem? Explain where he went wrong.

prohojiy [21]

Answer:

The slope is 5/-2.

Step-by-step explanation:

Slope is y1-y2 over m1-m2 (rise over run). The first ordered pair is -2 (m1) and 11 (y1). We then subtract the second ordered pair (4 (m2) and -4 (y2)) from the first.

11 - (-4) = 11 + 4 = 15

-2 - 4 = -6

Remember, slope is rise over run (y over x), so the slope is 15/-6. Now, we must simplify. 15/-6 = 5/-2

Dean went wrong because he thought that slope was run over rise (x over y). If he had switched the two numbers, his answer would have been correct.

8 0

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

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

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

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