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Gnoma [55]
3 years ago
13

Y=5x-2 is the equation of a straight line graph. Why is its gradient ?

Mathematics
1 answer:
olga55 [171]3 years ago
5 0
The gradient is the direction of steepest ascent. 5 is the coefficient of x, <span>y=mx+c,</span> so 5 is the gradient to this algebraic equation.<span />
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The work of a student to find the dimensions of a rectangle of area 8 + 12x and width 4 is shown below:
iogann1982 [59]

Answer:

Step 2: 4(2) + 4(3x)

Step-by-step explanation:

By definition, the area of a rectangle is given by:

Area = Length × Width

In this case, we know that:

Area = 8 + 12x

Width = 4

Therefore:

Step 1: 8 + 12x

Step 2: 4(2) + (4)(3x)

Step 3: 4(2 + 3x)

Therefore, the dimensions of the rectangle are 4 and 2 + 3x.

The mistake was made in STEP 2. Instead of 4(2) + 4(x2) it should be  4(2) + 4(3x). Which is the second option.

5 0
3 years ago
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expeople1 [14]
3/3 or it could be 1
4 0
3 years ago
A sphere with center A is shown. Which represents a tangent?<br> | WE
lana [24]

Answer:

<h2>The segment FE is a tangent of the sphere.</h2>

Step-by-step explanation:

Obseve the sphere in the image attached.

Notice that point A is the center of the sphere. GD is the diameter, which means AG and AD are radius.

On the other hand, remember that a tangent is a line that intersects the figure at only one point, otherwise, it would be a secant.

Therefore, segment BC is a secant and segment FE is tangent.

So, the right answer is segment FE.

4 0
3 years ago
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mezya [45]

Answer:

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Step-by-step explanation:

7 0
3 years ago
Find an equation, or a set of equations, to describe the set of points that are equidistant from the points p(−9, 0, 0) and q(3,
wariber [46]
X = -3.  
The distance from p(-9, 0, 0) is
 d = sqrt((x+9)^2 + y^2 + z^2) 
 The distance from q(3,0,0) is
 d = sqrt((x-3)^2 + y^2 + z^2) 
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 sqrt((x+9)^2 + y^2 + z^2) = sqrt((x-3)^2 + y^2 + z^2)
  Square both sides, then simplify
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 18x + 81 = - 6x + 9
 24x + 81 = 9
 24x = -72
 x = -3 
 So the desired equation is x = -3 which defines a plane.
7 0
3 years ago
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