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WARRIOR [948]
3 years ago
7

Keisha says that all functions are relations but not all relations are functions. kevin says that all relations are functions bu

t not all functions are relations. who is correct and why?
Mathematics
1 answer:
vladimir1956 [14]3 years ago
3 0

Keisha is correct, because as per the definition <u>A function is a special relationship where each input has a single output</u>.

A function is a special relation. In other words, a relation if and only if it has a specific characteristic where each input has a single output, then it is called a Function.

All functions are relations but not all relations are functions.


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What are these questions i need help
Over [174]

The roots of the equation f(x) = x^2 - 93987 are x = √93987 and x = -√93987

<h3>What are quadratic equations?</h3>

Quadratic equations are equations that have a second degree and have the standard form of ax^2 + bx + c = 0, where a, b and c are constants and the variable a does not equal 0

<h3>How to determine the other roots of the equation?</h3>

The equation of the function is given as:

f(x) = x^2 - 93987

The above equation is a quadratic equation

Express the equation as a difference of two squares

f(x) = (x - √93987)(x + √93987)

Set the equation of the function to 0

(x - √93987)(x + √93987) = 0

Split the factors of the above function equation as follows

x - √93987 = 0 and x + √93987 = 0

Solve for x in the above equations

x = √93987 and x = -√93987

Hence, the roots of the equation f(x) = x^2 - 93987 are x = √93987 and x = -√93987

Read more about roots of equation at

brainly.com/question/776122

#SPJ1

5 0
1 year ago
Which is the equation of a hyperbola with directrices at x = ±3 and foci at (4, 0) and (−4, 0)?​
Alex Ar [27]

Answer:

\frac{x^2}{12}-\frac{y^2}{4}=1

Step-by-step explanation:

Since our foci are located on the x-axis, then our major axis is going to be the horizontal transverse axis of the hyperbola:

<u>Formula for hyperbola with horizontal transverse axis centered at origin</u>

  • <u />\frac{x^2}{a^2}-\frac{y^2}{b^2}=1
  • Directrices -> x=\pm\frac{a^2}{c}
  • Foci -> (\pm c,0) where a^2+b^2=c^2
  • a>b

Since we are given our directrices of x=\pm3 and foci of (\pm4,0), then we can set up the directrices equation to solve for a^2:

x=\pm\frac{a^2}{c}\\ \\\pm3=\pm\frac{a^2}{4}\\ \\12=a^2

Now we can determine b^2 to complete our equation for the hyperbola:

a^2+b^2=c^2\\\\12+b^2=4^2\\\\12+b^2=16\\\\b^2=4

Therefore, our equation for our hyperbola is \frac{x^2}{12}-\frac{y^2}{4}=1

5 0
2 years ago
Need help on this please
Naily [24]
There is no outlier. :)
6 0
2 years ago
Read 2 more answers
Which pair of polygons is congruent?
erik [133]

Answer:

C

Step-by-step explanation:

Polygon 3 and 5 are congruent cause they have the same length of side

4 0
2 years ago
I need help plsssss...........
ch4aika [34]

Answer:

<h2>A)x=3/2 y=½</h2>

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>
  • system of linear equation
  • PEMDAS
<h3>let's solve:</h3>
  • since y is equal to both equation therefore we can substitute the value of y into the other equation
  1. \sf substitute \: the \: value \: of \: y :  \\  \sf  - x + 2 = 3x - 4
  2. \sf add \: x \: to \: both \: sides :  \\  \sf  - x + x + 2 = 3x + x - 4 \\  \sf 4x - 4 = 2
  3. \sf add \: 4 \: to \: both \: sides :  \\  \sf xx - 4 + 4 = 2 + 4 \\ 4x = 6
  4. \sf divide \: both \: sides \: by \: 4 :  \\  \frac{4x}{4}  =  \frac{6}{4}  \\ x =  \frac{3}{2}

let's figure out y

  1. substitute the got value of x into the second equation: y=3*3/2 -4
  2. simplify multiplication:9/2 -4
  3. simplify division:4.5-4
  4. substract:0.5 alternate form:y=½

3 0
2 years ago
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