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algol13
2 years ago
14

Help asap Use the graph to find the indicated value of the function.

Mathematics
2 answers:
mel-nik [20]2 years ago
8 0

If a pair related to the function is graphed

Then

  • (x,y) be a pair

So

f(x)=y

  • x is domain
  • y is range

Her e

  • f(-2)

Find the point

  • (-2,2)

So

f(-2)=2

kipiarov [429]2 years ago
7 0

f(-2) means the x value is -2 and you need to find the y value at that point.


At x = -2 the line crosses the y axis at y = 2

f(-2) = 2

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In the standard x,y plane, a circle has a radius 6 and center (7, 3). The circle intersects the x-axis at (a, 0) and (b, 0). Wha
arsen [322]

Answer:  The required value of a+b is 14.

Step-by-step explanation:  Given that in the standard xy plane, a circle has a radius 6 and center (7, 3).

The circle intersects the x-axis at (a, 0) and (b, 0).

We are to find the value of a+b.

We know that

the standard equation of a circle with center at (h, k) and radius r units is given by

(x-h)^2+(y-k)^2=r^2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

For the given circle, we have

(h, k) = (7, 3)  and  r = 6 units.

So, from equation (i), we get

(x-7)^2+(y-3)^2=6^2\\\\\Rightarrow (x-7)^2+(y-3)^2=36~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)

Since the circle (ii) passes through the points (a, 0) and (b, 0), so let the point be denoted by (c, 0), then we have

(c-7)^2+(0-3)^2=36\\\\\Rightarrow (c-7)^2+9=36\\\\\Rightarrow (c-7)^2=27\\\\\Rightarrow c-7=\pm3\sqrt3~~~~~~~~~~~~~~~~~~~~~~[\textup{Taking square root on both sides}]\\\\\Rightarrow c=7\pm3\sqrt3

Therefore, we get

a=7+3\sqrt3,\\\\b=7-3\sqrt3.

That is,

a+b=(7+3\sqrt3)+(7-3\sqrt3)=14.

Thus, the required value of a+b is 14.

4 0
3 years ago
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Tatiana [17]

For this case we have the following expression:

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By definition, a constant term is a value that cannot be modified, does not vary.

It is observed that the first three terms of the expression depend on a variable, therefore they are not constant.

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The constant term is 3

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Answer:

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