Are the triangles AA, SAS, SSS or not similar ?

) If the gradient of ff is ∇f=xi⃗ +yj⃗ +3zyk⃗ ∇f=xi→+yj→+3zyk→ and the point P=(−6,−6,−2)P=(−6,−6,−2) lies on the level surface

Paha777 [63]

Answer:

-3x-3y-z=38

Step-by-step explanation:

To find the equation of the tangent plane you can use

$\bigtriangledown f_x(x_0,y_0,z_0)(x-x_0)+\bigtriangledown f_y(x_0,y_0,z_0)(y-y_0)+\bigtriangledown f_z(x_0,y_0,z_0)(z-z_0)=0$

the point P is (-6,-6,-2). Hence you have

$(-6)(x-(-6))+(-6)(y-(-6))+(-2)(z-(-2))=0\\-6x-36-6y-36-2z-4=0\\-6x-6y-2z=76\\-3x-3y-z=38$

hope this helps!!

5 0

3 years ago

Help plz i dont undertand

lora16 [44]

Answers:

x = 4
AB = 36
BC = 36
AC = 36

All three sides are 36 units long.

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Explanation:

Triangle ABC is equilateral, so all three sides are the same length.

Pick any two of those sides, set them equal to each other, and solve for x

AB = BC

8x+4 = 11x-8

8x-11x = -8-4

-3x = -12

x = -12/(-3)

x = 4

Use this x value to find each side. We should get the same value (if not, then an error occurred in solving for x).

AB = 8x+4 = 8*4+4 = 32+4 = 36
BC = 11x-8 = 11*4-8 = 44-8 = 36
AC = 9x = 9*4 = 36

We get the same value for the three side lengths, so this confirms we have the correct x value.

7 0

2 years ago

A billboard designer has decided that a sign should have 3 ft margins at the top and bottom and 5 ft margins on the left and rig

elixir [45]

Answer:

The width of billboard is "[x]" and the height of billboard is "[y"]. If total area of billboard is $9000 ft^2$ then $9000=xy$

Step-by-step explanation:

• The total width of billboard is [x]. Therefore the width of printed area will be (x-10) by excluding margin of left and right side.

• The total height of billboard is [y]. Therefore the height of printed area will be [(y-6)] by excluding the margin of top and bottom from the total height.

• To find the printed area of billboard calculations are given below:

$& 9000=xy$

$& y=\frac{9000}{x} \\ & A=(x-10)(y-6) \\ & A=xy-6x-10y+60 \\ & A=x\left( \frac{9000}{x} \right)-6x-10\left( \frac{9000}{x} \right)+60 \\ & A=9060-6x-\frac{9000}{x} \\$

On taking the first order derivative of A

$A'=-6+\left( \frac{90000}{{{x}^{2}}} \right)$

$& \left( \frac{90000}{{{x}^{2}}} \right)-6=0 \\ & 6{{x}^{2}}=90000 \\ & x=\sqrt{15000} \\ & y=\frac{9000}{x}=\frac{90000}{\sqrt{15000}}=10\sqrt{150} \\$