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anygoal [31]
3 years ago
14

6. HELP W THIS ONE TOO PLS!!

Mathematics
1 answer:
dybincka [34]3 years ago
5 0
This occurs for the input(s) x where the outputs, f(x) and g(x), are equal. So here, f(x) = g(x) when x = 1, because 5 = 5.
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Which equation represents the parabola shown on the
EastWind [94]

Answer:

y2=-8x is what's graphed so B.

Step-by-step explanation:

(I think you mistyped the choice)

3 0
3 years ago
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The lateral area of a right cylinder which has a base diameter of 12 cm and a height of 8 cm is?
garik1379 [7]

Answer:

A = 301.59\ cm^2 or A=96\pi\ cm^2

Step-by-step explanation:

The lateral area of a cylinder is calculated by the following formula

A = 2\pi r * h.

Where r is the radius of the right cylinder and h is the height

In this case we know that the diameter d of the cylinder is

d=2r\\\\r=\frac{d}{2}\\\\r=\frac{12}{2}\\\\r=6\ cm

h=8\ cm

Therefore the lateral area is:

A = 2\pi*6 * 8.

A = 96\pi

A = 301.59\ cm^2

8 0
3 years ago
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A polynomial P(x) and a divisor d(x) are given. Use long division to find a quotient Q(x) and the remainder R(x). Express P(x) i
Vilka [71]
It's a bit hard to explain but I hope this helps

5 0
3 years ago
the measure of angle is 7π/6. the measure of its reference angle is *blank* degrees, and sin is *blank*.
marin [14]
The measure of angle θ = 7pi/6. To convert this to degrees, substitute pi with 180 degrees. Pi is always equal to 180 degrees.
θ becomes: 7 x 180/6
Therefore, θ = 210 degrees.
And, to solve for sin θ, you can directly input sin (210) in the calculator. 
Thus, sin θ = sin (210) = -0.5 or -1/2
7 0
3 years ago
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Find the equation for the plane that contains the line x=−1+3t , y=1+2t, z=2+4t and is perpendicular to the plane containing the
Ivan

Let L be the line given by the vector equation

(-1,1,2) + \lambda(3,2,4) \ , \lambda \in \mathbb{R}.

First, we use the director vectors of the lines L1 and L2 to get the

vector equation of the plane containing them, which we denote by \Pi_1. This is,

\\\\\Pi_1  : (1,-1,5) + \alpha (2,-1,6) + \beta (1,1,-3) \ , \alpha, \beta \in \mathbb{R}\\\\\\

We observe that \vec{N} = (2,-1,6)\times(1,1,-3) = (-3,12,3) \ne (0,0,0). Therefore, the vector equation of \Pi_1 defines a plane and \vec{N} is a normal vector to \Pi_1.

 

Finally, the vector equation for the wanted plane, which we denote by \Pi, is

\Pi : (-1,1,2) + r(3,2,4) + s(-3,12,3), r,s \in \mathbb{R} \ .

Thus, if s = 0, then L \subset \Pi and since \vec{N} is parallel to \Pi, then it is perpendicular to \Pi_1.

8 0
3 years ago
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