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4 years ago

Which relation represents a function

Mathematics

1 answer:

BARSIC [14]4 years ago

3 0

Answer:

A relation is a function if each element of the domain is paired with exactly one element of the range. If given a graph, this means that it must pass the vertical line test.

Step-by-step explanation:

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Write a rule for finding the value of any base with an exponet of 0

Julli [10]

The exponent of 0 makes everything equal to 1

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4 years ago

The drawing below represents a 22-foot ladder with its base placed 10 feet from the bottom of a vertical wall.

Nesterboy [21]

Answer:

63°

Step-by-step explanation:

To get the value of x we will use the Pythagoras theorem

hyp² = opp²+adj²

Given

hyp is the length of the ladder = 22ft

Adjacent is the distance from the base of the ladder to the wall = 10ft

Angle of elevation = x

Using the SOH CAH TOA identity

Cos theta = adjacent/hypotenuse

Cos x = 10/22

Cos x = 5/11

x = arccos(5/11)

x = arccos(0.4545)

x = 62.96°

Hence the best estimate of x is 63°

4 0

3 years ago

Express (x – 3)2 as a trinomial in standard form.

Sindrei [870]

Answer:

x^2 - 6x + 9

Step-by-step explanation:

5 0

3 years ago

What is the distance between P(3,-2) and Q(-5,-2) ?

Liono4ka [1.6K]

Answer:

Step-by-step explanation:

8 0

3 years ago

What causes a solution to a rational equation to be an extraneous solution?

andrey2020 [161]

Extraneous solution:

An extraneous solution is a solution that arises from the solving process that is not really a solution at all

So, firstly we will solve for the equation

and then we verify each solutions by plugging them back

If denominator of rational equation becomes zero , then that solution must be extraneuous solution

For example:

$\frac{1}{x+3} +\frac{2}{x} =-\frac{3}{x(x+3)}$

we can solve for x

Multiply both sides by x(x+3)

$\frac{1}{x+3}x\left(x+3\right)+\frac{2}{x}x\left(x+3\right)=-\frac{3}{x\left(x+3\right)}x\left(x+3\right)$

now, we can simplify it

$x+2\left(x+3\right)=-3$

$x+2x+6=-3$

$3x=-9$

now, we can solve for x

$x=-3$

now, we can check whether x=-3 is extraneous solution

we will plug back x=-3 into original

$\frac{1}{-3+3} +\frac{2}{-3} =-\frac{3}{-3(-3+3)}$

$\frac{1}{0} +\frac{2}{-3} =-\frac{3}{-3(0)}$

we can see that denominator becomes 0

so, x=-3 can not be solution

so, x=-3 is extraneous solution...........Answer

6 0

3 years ago