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3 years ago

Does the table represent a function?

Mathematics

2 answers:

asambeis [7]3 years ago

4 0

Yes, the table represents a function.

None of the independent (x) values are repeated and each one has a corresponding (y) value.

If you had a repeated (x) value in the table, it would not represent a function.

Alinara [238K]3 years ago

4 0

Answer:

The table represents a function

Step-by-step explanation:

a function is a map from non-empty set A to a non-empty set B such that every element of A is mapped to a unique element of B.

As we can see from the table that corresponding to every x, there exist a unique image y

(f(-2)=7

f(-1)=4

f(0)=3

f(1)=4

f(2)=7)

Hence, The table represents a function.

You might be interested in

Can anyone please help me :)

Solnce55 [7]

Answer:

5 to the 15th power.

Step-by-step explanation:

PEMDAS requires us to do parenthesis, then exponents, and then division. Let's do 5³. 5x5x5 is equal to 125. Now, we find 125 to the ninth power. According to my calculations, 125 to the ninth power is equal to 7,450,580,596,923,828,125. Now, we find 5 to the twelfth power, which, according to my calculations, is 244,140,625. Now, we solve. According to my calculations, 7,450,580,596,923,828,125/244,140,625 is equal to 30,517,578,125. 30,517,578,125 is equivalent to 5 to the 15th power. That means that the second option is correct. I hope this helps!

3 0

3 years ago

Which one of the properties described below DOES NOT apply to the perpendicular bisector of a segment?

Lana71 [14]

Answer:

It must be longer than the original segment.

Step-by-step explanation:

Let analyse all the 4 answers:

a. It divides the original segment into two equal pieces.

Yes because It finds the midpoint of the given line segment.

b. Every point on the perpendicular bisector is the same distance from both end points of the segment.

Yes, because perpendicular bisector theorem states that if a point is on the perpendicular bisector of a segment, then it is equidistant from the segment's both endpoints.

c. It must be longer than the original segment

Wrong, it can be shorter or equal to the original segment.

d. It is perpendicular to (makes a 90 angle with) the original segment

True, because of it is one of the the properties of a perpendicular bisector of a segment

So we choose C.

Hope it will find you well.

7 0

3 years ago

Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.

gavmur [86]

The area of the part of the plane 3x 2y z = 6 that lies in the first octant is mathematically given as

A=3 √(4) units ^2

<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

$A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)$

The partial derivatives of a function are f x and f y.

$\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}$

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

$&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\$

$&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}$

In conclusion, the area is

A=3 √4 units ^2

Read more about the plane

brainly.com/question/1962726

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5 0

1 year ago

Which is the equation of a circle that has a center of (-1, 3) and passes through the point (3, 6)?

tatiyna

Answer:

( x + 1)^2 + ( y - 3)^2 = 25

Step-by-step explanation:

The equation of the circle with a center and a point

( x - a) ^2 + ( y - b) ^2 = r^2

( a , b) - center of the circle

( x , y) - any point on the circle

r^2 - radius

( -1 , 3) - ( center) - ( a, b)

a = -1

b = 3

( 3 , 6) - ( point) - ( x, y)

x = 3

y = 6

Step 1: substitute the center into the equation

( x -(-1)^2 + ( y - 3)^2 = r^2

( x + 1)^2 + ( y - 3)^2 = r^2

Step 2: sub the point into the equation

( x + 1)^2 + (y - 3)^2 = r^2

( 3 + 1)^2 + ( 6 - 3)^2 = r^2

4^2 + 3^2 = r^2

16 + 9 = r^2

25 = r^2

Step 3: sub the radius into the equation

( x + 1)^2 + ( y - 3)^2 = r^2

( x + 1)^2 + (y - 3)^2 = 25

Therefore, the equation of the circle is

( x + 1)^2 + (y - 3)^2 = 25

6 0

3 years ago

Write a second inequality withe the same meaning -20 <- s

DedPeter [7]

S < 20 , -5x > -100 , etc

8 0

3 years ago