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2 years ago

What is the simplest form of this problem?

Mathematics

2 answers:

Georgia [21]2 years ago

7 0

The answer is 21 rootunder 15

JulijaS [17]2 years ago

3 0

Answer:

21 rootunder 15

Step-by-step explanation:

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5. If "h" is positive how does the parabola move? If negative?

Paul [167]

Answer: positive values for h translate the parabola to the right, negative values for h translate the parabola to the left

Step-by-step explanation: Note that the value for h is subtracted from x, and then this difference is squared. See that positive values for h translate the parabola to the right, negative values for h translate the parabola to the left

7 0

2 years ago

1/3into a percent and a decimal

jeka57 [31]

Answer:

33% or .33

Step-by-step explanation:

Divide 1/3 = .3333 (repeating). I would round to the nearest hundreths place:

1/3 = .33

For the precentage, multiple 0.33 by 100:

100 * 0.33 = 33%

3 0

2 years ago

MATHHHHHHHHHHHHHHHHHHHHHHHHHH

Svetradugi [14.3K]

Answer:

x intercept = 3

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3 0

3 years ago

Read 2 more answers

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

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

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

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