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3 years ago

given the graph function below which of the following ordered pairs are found on the inverse function

Mathematics

1 answer:

xenn [34]3 years ago

4 0

Answer:

Option A.

Step-by-step explanation:

Graph of function has been given in the figure.

Coordinates of the points lying on this graph will be

x -2 -1 0 1 2

y -10 -3 -2 -1 6

Therefore, coordinates of the points which will lie on the inverse of this function will be

x -10 -3 -2 -1 6

y -2 -1 0 1 2

Therefore, Option A. will be the answer.

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Solve for a side in right triangles

natita [175]

Answer:

2.30

Step-by-step explanation:

The side of interest is opposite the given angle, and the hypotenuse is given. The sine relation tells you ...

sin(A) = a/c

a = c·sin(A) = 3·sin(50°)

a = BC ≈ 2.2981 ≈ 2.30

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4 years ago

HELP ASAP!!!!! Given: Point B is on the perpendicular bisector of AC¯¯¯¯¯. BD¯¯¯¯¯ bisects AC¯¯¯¯¯ at point D. Prove: B is equid

Harlamova29_29 [7]

Answer: Missing parts are,

In first blank, $AD\cong DC$ ,

In second blank, SAS postulate

In third blank, CPCTC postulate

Step-by-step explanation:

Since, Here D is the mid point on the line segment AC.

And BD is a perpendicular to the line AC.

Therefore, In triangles ADB and CDB ( shown in figure)

AD\cong DC ( By the definition of mid point)

$\angle BDA\cong \angle BDC$ ( right angles )

$BD\cong BD$ ( reflexive)

Thus, By SAS ( side angle side )postulate,

$\triangle ADB\cong \triangle CDB$

So, by CPCTC( Corresponding parts of congruent triangles are congruent)

$AB\cong CB$

Now, By definition of congruent segment,

AB=CB

By definition of equidistant,

B is equally far from both A and C.

7 0

3 years ago

Read 2 more answers

What does pie equal if its equal

katen-ka-za [31]

Trick question Pie can never be equal Pie is Unlimted

6 0

3 years ago

A. x=45°,y=13.1x=45°,y=13.1

Kryger [21]

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3 years ago

Which of the following is equivalent to 6x 2 + x - 12 ?

bagirrra123 [75]

$6x^2+x-12=6x^2-9x+8x-12=3x(2x-3)+4(2x-3)\\\\=(2x-3)(3x+4)\\\\other\ method:\\\\6x^2+x-12\\a=6;\ b=1;\ c=-12\\\Delta=b^2-4ac\to\Delta=1^2-4\cdot6\cdot(-12)=1+288=289\\\\x_1=\frac{-b-\sqrt\Delta}{2a};\ x_2=\frac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{289}=17$

$x_1=\frac{-1-17}{2\cdot6}=\frac{-18}{12}=-\frac{3}{2};\ x_2=\frac{-1+17}{2\cdot6}=\frac{16}{12}=\frac{4}{3}\\\\6x^2+x-12=6(x+\frac{3}{2})(x-\frac{4}{3})=2\cdot3(x+\frac{3}{2})(x-\frac{4}{3})\\\\=2(x+\frac{3}{2})\cdot3(x-\frac{4}{3})=(2x+3)(3x-4)$

4 0

3 years ago