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3 years ago

PICKING BRIANLIEST FOR THE CORRECT ANSWER!!!!!!!!!!!

Mathematics

2 answers:

irga5000 [103]3 years ago

7 0

Im pretty sure you put the wrong picture. The slope isn't negative and there is no point O,A, or B.

Veseljchak [2.6K]3 years ago

4 0

Where is O, A and B?

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A store called Surfmania sells a $170 surfboard for 30% off. Another store, Coolwaves, sells the same surfboard but for $150, bu

Marizza181 [45]

Answer:

no se467/'%7&!&?!&$-!---;!8

6 0

3 years ago

Let φ(x, y) = arctan (y/x) .

Alexandra [31]

Answer:

a) $\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})$

b) $\large \mathbb{R}^2-\{(0,0)\}$

c) the points of the form (x, -x) for x≠0

Step-by-step explanation:

If φ(x, y) = arctan (y/x), the vector field F = ∇φ would be

$\large F(x,y)=(\frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial y})$

On one hand we have,

$\large \frac{\partial \phi}{\partial x}=\frac{\partial arctan(y/x)}{\partial x}=\frac{-y/x^2}{1+(y/x)^2}=-\frac{y/x^2}{1+y^2/x^2}=\\\\=-\frac{y/x^2}{(x^2+y^2)/x^2}=-\frac{y}{x^2+y^2}$

On the other hand,

$\large \frac{\partial \phi}{\partial y}=\frac{\partial arctan(y/x)}{\partial y}=\frac{1/x}{1+(y/x)^2}=\frac{1/x}{1+y^2/x^2}=\\\\=\frac{1/x}{(x^2+y^2)/x^2}=\frac{x}{x^2+y^2}$

$\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})$

The domain of definition of F is

$\large \mathbb{R}^2-\{(0,0)\}$

i.e., all the plane X-Y except the (0,0)

Here we want to find all the points such that

$\large (-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})=(k,k)$

where k is a real number other than 0.

But this means

$\large -\frac{y}{x^2+y^2}=\frac{x}{x^2+y^2}\Rightarrow y=-x$

So, all the points in the line y = -x except (0,0) are parallel to the vector field F, that is, the points (x, -x) with x≠ 0

8 0

3 years ago

3. Given ABCD ≅ FGHJ and mC = 9x - 7, mH = 5x + 13, find the value of x and the measures of angle C and angle H. Show all work

Bas_tet [7]

Since congruence and similarity for polygons is posted in a specific order, we know that angle A=angle F, side A=side F, die B=side G, and so on. Therefore, we have that angle C=angle H = 9x-7=5x+13. Subtracting 5x from both sides to separate the x using the subtraction property of equality, we get 4x-7=13. Next, we can add 7 to both sides to get 4x=20, and divide both sides by 4 to isolate the x and get x=5. Plugging that into angle C or angle H, we get that 9x-7=9*5-7=45-7=38=5*5+13.

Feel free to ask further questions!

5 0

3 years ago

Need help on these !! 13 and 14! Please!!

katen-ka-za [31]

Question # 13

Answer:

The required equation for the given function is y = 4sin(x/2+2π/3) -2 , as shown attached graph diagram.

Step-by-step explanation: 

As the general sine function is given by

$y=asin(bx+c)+d$ .......[A]

amplitude = a
period = 2π ÷ b
Phase shift = -c ÷ b
Vertical shift = d

As in the question,

amplitude = a = 4
period = 4π
phase shift = -4π/3
Vertical shift = d = -2

period = 2π ÷ b

b = 2π/period

b = 2π/4π ∵ period = 4π

b = 1/2

Also

Phase shift = -c/b

-4π/3 = -c/b ∵ phase shift = -4π/3

4π/3 = c/b

c = b × 4π/3

c = 1/2 × 4π/3

c = 4π/6

c = 2π/3

So, putting Amplitude ⇒ a = 4, Vertical shift ⇒ d = -2, b = 1/2 ,

and c = 2π/3 in Equation [A] would bring us the required equation for the given function.

$y=asin(bx+c)+d$

y = 4sin(x/2+2π/3)+(-2)

y = 4sin(x/2+2π/3) -2

Note: The graph is also shown in attached diagram.

Question # 14

Answer:

The required equation for the given function is y = cot(x+π/3)+2, as shown in attached graph diagram.

Step-by-step explanation: 

As the general cotangent function is given by

$y=acot(bx+c)+d$ .......[A]

amplitude = a
period = π ÷ b
Phase shift = -c ÷ b
Vertical shift = d

As in the question,

period = π
phase shift = -π/3
Vertical shift = d = 2

period = π ÷ b

b = π/period

b = π/π ∵ period = 4π

b = 1

Also

Phase shift = -c/b

-π/3 = -c/b ∵ phase shift = -π/3

π/3 = c/b

c = b × π/3

c = 1 × π/3

c = π/3

So, putting vertical shift ⇒ d = 2, b = 1 and c = π/3 in Equation [A] would bring us the required equation for the given function.

$y=acot(bx+c)+d$

y = cot(x+π/3)+2

Note: The graph is also shown in attached diagram.

Keywords: amplitude, period , phase shift , vertical shift

Learn more about trigonometric functions of equations from brainly.com/question/2643311

#learnwithBrainly

7 0

4 years ago

Whats the bisector of the photosentisis of the yellow sun???!??!??!

LenKa [72]

Answer:

you will use dy DX for the equation of photosynthesis. make carbon subject of formular, and factorize the final answer

7 0

3 years ago