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3 years ago

1 (9x+2)=4+ 2 (6x+4) 3 3

Mathematics

1 answer:

mote1985 [20]3 years ago

6 0

Answer:i think

Step-by-step explanation:

1(9x+2)=4+2(6x+4)33

=9 x (396 x + 268) + 2 (396 x + 268)

=9 * 396 x x+2 * 396 x+9 * 268 x+2 * 268

=3564 x^2 + 3204 x + 536

(x^2+2)(x+8) (or) (x^3+5)(x+6) (or) (x+2)(x+8)(x+6)

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Solve ? -4-4(-x-1)=-4(6+2x)

soldier1979 [14.2K]

$-4-4(-x-1)=-4(6+2x)\\\\-4+4x+4=-24-8x\\\\4x=-24-8x \ \ /+8x\\\\12x=-24 \ \ /:12\\\\\huge\boxed{x=-2}$

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3 years ago

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

sergiy2304 [10]

Answer:

The measure of angle θ is 7π/6. The measure of its reference angle is 210°

and sin θ is -1/2.

Step-by-step explanation:

The correct question is:

The measure of angle θ is 7π/6. The measure of its reference angle is ___

and sin θ is ___

180° is equivalent to π radians. To transform 7π/6 radians to degrees, we have to use the following proportion:

180° / π radians = x° / (7π/6 radians)

x = (180/π) * (7π/6 radians)

x = 210°

And sin(210°) = -1/2

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3 years ago

Y = 1/2x +10 y = 4x - 4 solve by graphing

Firdavs [7]

Answer:

(4,12)

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3 years ago

Complete the graph below with 3 points plotted to earn 30 points and brainliest :))

Firdavs [7]

Answer:

Point 1: (-5.03, 0.001)

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Step-by-step explanation:

3 0

3 years ago

At what point on the paraboloid y = x2 + z2 is the tangent plane parallel to the plane 3x + 2y + 7z = 2? (if an answer does not

Nikitich [7]

If $f(x, y, z) = c$ represent a family of surfaces for different values of the constant $c$ . The gradient of the function $f$ defined as $\nabla f$ is a vector normal to the surface $f(x, y, z) = c$ .

Given the paraboloid

$y = x^2 + z^2$ .

We can rewrite it as a scalar value function f as follows:

$f(x,y,z)=x^2-y+z^2=0$

The normal to the paraboloid at any point is given by:

$\nabla f= i\frac{\partial}{\partial x}(x^2-y+z^2) - j\frac{\partial}{\partial y}(x^2-y+z^2) + k\frac{\partial}{\partial z}(x^2-y+z^2) \\ \\ =2xi-j+2zk$

Also, the normal to the given plane $3x + 2y + 7z = 2$ is given by:

$3i+2j+7k$

Equating the two normal vectors, we have:

$2x=3\Rightarrow x= \frac{3}{2} \\ \\ -1=2 \\ \\ 2z=7\Rightarrow z= \frac{7}{2}$

Since, -1 = 2 is not possible, therefore there exist no such point on the paraboloid $y = x^2 + z^2$ such that the tangent plane is parallel to the plane 3x + 2y + 7z = 2.

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3 years ago