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zzz [600]
3 years ago
15

For this graph, mark the statements that are true.

Mathematics
2 answers:
IRISSAK [1]3 years ago
4 0
Hello,

I was asking if i may reply as genius are posting so much questions?
(Is-it in order to make the schmilblick(if you know what it is) go on?)

y=f(x)=(x-1)²

A is True Dom f(x)=R

B is false there are no negative values of y

C is false since A is true.

D is true IMG(f(x))=R+.


Vitek1552 [10]3 years ago
4 0

Answer:

Options A and D

Step-by-step explanation:

The given graph for a parabola can be represented by the equation

y = (x - h)² + k

where (h, k) is the vertex of the parabola.

Since vertex of the parabola is (1, 0) so the equation will be

y = (x - 2)² + 0

y = x² -4x + 4

Since domain of the parabola is defined by its x-values

Therefore, domain will be set of all real numbers

Now we know range of a function on graph is defined by the y-values.

Therefore, Range of the given function will be set of all real numbers greater than of equal to zero.

Options A and D are the correct options.

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F(x) = x3 -27<br> Find the zeros
lara [203]

Answer:

x=9

y=-27

Step-by-step explanation:

6 0
3 years ago
3. (6 points) Determine whether the relation R on the set A is an equivalence relation a. (3 pts) A = {1,2,3,4, 5) R={(1,1), (1,
dezoksy [38]

Answer:

a is not an equivalence relation.

b is an equivalence relation.

Step-by-step explanation:

a.

A = {1,2,3,4, 5) R={(1,1), (1,2), (1,3), (2,2), (2,3), (3,1), (3,2), (3,3), (4,4), (5,5)

To see if is an equivalence relation you need to see if you have these 3 things:

Part 1: xRx for all x in A. This is the reflexive property.

Do we? Yes we have all these points in R: (1,1), (2,2) ,(3,3) ,(4,4), and (5,5).

Part 2: If xRy then yRx. This is the symmetic property.

Do we? We have (1,2) but not (2,1). So it isn't symmetric.

Part 3: If xRy and yRz then xRz.

Do we? We are not going to check this because there is no point. We have to have all 3 parts fot it be an equivalence relation.

b.

A = {a, b, c} R={(a, a), (a, c), (b, b), (c, a), (c, c)}

To see if is an equivalence relation you need to see if you have these 3 things:

Part 1: xRx for all x in A. This is the reflexive property.

Do we? Yes we have all these points in R: (a,a),(b,b), and (c,c).

Part 2: If xRy then yRx. This is the symmetric property.

Do we? We have (a,c) and (c,a). We don't need to worry about any other (x,y) since there are no more with x and y being different. This is symmetric.

Part 3: If xRy and yRz then xRz.

Do we? We do have (a,c), (c,a), and (a,a).

We do have (c,a), (a,c), and (c,c).

So it is transitive.

Question b has all 3 parts so it is an equivalence relation.

4 0
3 years ago
Which phrase should be inserted on the line to make a true statement
Oxana [17]
They’re equal. 8 is 2/3 of 12.
4 0
3 years ago
Read 2 more answers
Does anyone know the answer pleaseeee
pochemuha

Answer:

it b

Step-by-step explanation:

3 0
3 years ago
" A car is travelling around a horizontal circular track with radius r = 230 m at a constant speed v = 17 m/s as shown. The angl
Luda [366]
R = 230 m
v = 17 m/s

Formula: |Ac| = v^2 / r

Point A

|Ac| = v^2 / r = (17m/s)^2 / (230m) = 1.257 m/s^2

Components:

cos(25) = - [ x-component / |Ac| ] => x-component = -|Ac|*cos(25)
sin(25)  = - [y-component / |Ac| ] => y-component = - |Ac|*sin(25)

x-component = - 1.256 * cos(25) m/s^2 = - 1.139 m/s^2
y-component =  -1.256 * sin (25) m/s^2 = - 0.531 m/s^2

Point B

|Ac| = 1.256

x-component = - 1.256 cos(58) = - 0.666 m/s^2
y-component = + 1.256 sin(58) = + 1.065 m/s^2

 

 
7 0
3 years ago
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