Does anyone know how to solve one of these ?

Given right triangle ABC, what is the value of tan(A)?

Angelina_Jolie [31]

You didn't provide more information regarding triangle but from given options, it seems that triangle has ratio from shortest side to longest side would be 5 : 12 : 13

So 13 would be hypotenuse since it is longest.

Now A cannot be right angle because tangent is not defined at 90°.

Also tan(A) would be ratio of opponent side to adjacent side (think SOH CAH TOA)

So either opposite side would be 5 or 12, and adjacent side would be 12 or 5 respectively.

So the answer is either 5 / 12 or 12 / 5 but only option provided is 12/5 so that would be your answer.

Final answer: 12 / 5

7 0

3 years ago

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One

lesantik [10]

$\bf \begin{array}{ccll} oz&calories\\ \cline{1-2} 6&90\\ 64&x \end{array}\implies \cfrac{6}{64}=\cfrac{90}{x}\implies \cfrac{3}{32}=\cfrac{90}{x}\implies 3x=2880 \\\\\\ x=\cfrac{2880}{3}\implies x=960$

3 0

2 years ago

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1. 48 c = __qt

Scorpion4ik [409]

Number 1:
We know that 1 quart = 4 cups. That means we need to divide 48 by 4.
What is 48 / 4=12
Number 1 is D
Number 2
We know that oz =0.0625 pounds
Well we multiply 0.0625 by 92 and get 6
Number 2 is B
Number 3
We know that 1 yd = 3 ft
We have to multiply. 3*6=18 But we have to add that one 18+1=19
Number 3 is C
Number 4
We know that 6 pounds = 96 oz from earlier and half of a pound is 8 so that means we do 96+8=104
Number 4 is A

Hope this really helped!

5 0

2 years ago

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A student was asked to find the equation of the tangent plane to the surface z=x3−y4 at the point (x,y)=(1,3). The student's ans

aliya0001 [1]

Answer:

C) The partial derivatives were not evaluated a the point.

D) The answer is not a linear function.

The correct equation for the tangent plane is $z = 241 + 3 x - 108 y$ or $3x-108y-z+241 = 0$

Step-by-step explanation:

The equation of the tangent plane to a surface given by the function $S=f(x, y)$ in a given point $(x_0, y_0, z_0)$ can be obtained using:

$z-z_0=f_x(x_0,y_0,z_0)(x-x_0)+f_y(x_0,y_0, z_0)(y-y_0)$ (1)

where $f_x(x_0, y_0, z_0)$ and $f_y(x_0, y_0, z_0)$ are the partial derivatives of $f(x,y)$ with respect to $x$ and $y$ respectively and evaluated at the point $(x_0, y_0, z_0)$ .

Therefore we need to find two missing inputs in our problem in order to use equation (1). The $z_0$ coordinate and the partial derivatives $f_x(x_0, y_0, z_0)$ and $f_y(x_0, y_0, z_0)$ . For $z_0$ just evaluating in the given function we obtain $z_0= -80$ and the partial derivatives are:

$\frac{\partial f(x,y)}{\partial x} \equiv f_x(x, y)= 3x^2 \\f_x(x_0, y_0) = f_x(1, 3) = 3$

$\frac{\partial f(x,y)}{\partial y} \equiv f_y(x, y)= -4y^3 \\f_y(x_0, y_0) = f_y(1, 3) = -108$

Now, substituting in (1)

$z-z_0=f_x(x_0,y_0,z_0)(x-x_0)+f_y(x_0,y_0, z_0)(y-y_0)\\z + 80 = 3x^2(x-1) - 4y^3(y-3)\\z = -80 + 3x^2(x-1) - 4y^3(y-3)$

Notice that until this point, we obtain the same equation as the student, however, we have not evaluated the partial derivatives and therefore this is not the equation of the plane and this is not a linear function because it contains the terms ( $x^3$ and $y^4$ )

For finding the right equation of the tangent plane, let's substitute the values of the partial derivatives evaluated at the given point:

$z = -80 + 3x^2(x-1) - 4y^3(y-3)\\z = -80 +3(x-1)-108(y-3)\\z = 241 + 3 x - 108 y$

or $3x-108y-z+241 = 0$

7 0

3 years ago

What is the solution to the system?

musickatia [10]

Just you need graphing these two lines and where will intersect these two lines so this will be the solution of the system

8 0

2 years ago