All categories

4 years ago

G = {(0, 1)} Is G-1 a function and why?

Mathematics

2 answers:

frosja888 [35]4 years ago

7 0

Answer:

Yes, each element in the domain has only one range value.

Step-by-step explanation:

I know because I just took a test and got it right. ;)

Liono4ka [1.6K]4 years ago

5 0

Answer:

From the information provided you cannot conclude whether or not G-1 is a function. You would need more coordinates in the set to conclude this answer.

You might be interested in

Which sequence of transformations on the red triangle will map it onto the missing portion of the square?

anyanavicka [17]

Answer:

A 90° counterclockwise rotation about the origin and then a translation 16 units right and 16 units up

Solution -Rotating the triangle 90° counterclockwise will take the triangle to 3rd quadrant and then further moving it 16 steps right will take it to 4th quadrant and followed by 16 steps upward will take it to the desired position which is in 1st quadrant.

7 0

3 years ago

8x+3=8x I need help

ANEK [815]

X=5/8 that is the answer lol

5 0

3 years ago

Look at the factors of 24 and 32.

Salsk061 [2.6K]

Answer:

Hope this helps

4 0

2 years ago

Read 2 more answers

Helpppp assaaaaaaap <br>determine the equation of line (AB)<br>A(I;1) and B(0;5)

prohojiy [21]

Y = -4x +5 should be right!

6 0

3 years ago

In the trapezoid ABCD (AB∥CD) point M∈AD, so that AM:MD=3:5. Line L ∥AB and going through point M intersects diagonal AC and leg

siniylev [52]

Answer:

$\dfrac{AP}{PC}=\dfrac{3}{5}$

$\dfrac{BN}{CN}=\dfrac{3}{5}$

Step-by-step explanation:

Consider triangles AMP and ADC. In these triangles,

angle A is the common angle, so $\angle MAP\cong \angle DAC$ by reflexive property;
angles AMP and ADC are congruent as corresponding angles when two parallel lines MP and CD are cut by transversal AD.

Hence, triangles AMP and ADC are similar by AA similarity theorem.

Similar triangles have proportional corresponding sides, thus

$\dfrac{AM}{AD}=\dfrac{AP}{AC}\\ \\\dfrac{3x}{3x+5x}=\dfrac{AP}{AC}\\ \\\dfrac{AP}{AC}=\dfrac{3}{8}\Rightarrow AP=\dfrac{3}{8}AC\\ \\PC=AC-AP=AC-\dfrac{3}{8}AC=\dfrac{5}{8}AC,$

$\dfrac{AP}{PC}=\dfrac{\frac{3}{8}AC}{\frac{5}{8}AC}=\dfrac{3}{5}$

Consider triangles ACB and PCN. In these triangles,

angle C is the common angle, so $\angle ACB\cong \angle PCN$ by reflexive property;
angles ABC and PCN are congruent as corresponding angles when two parallel lines PN and AB are cut by transversal BC.

Hence, triangles ACB and PCN are similar by AA similarity theorem.

Similar triangles have proportional corresponding sides, thus

$\dfrac{CP}{AP}=\dfrac{CN}{CB}\\ \\\dfrac{5x}{3x+5x}=\dfrac{CN}{CB}\\ \\\dfrac{CN}{CB}=\dfrac{5}{8}\Rightarrow CN=\dfrac{5}{8}CB\\ \\BN=BC-CN=BC-\dfrac{5}{8}BC=\dfrac{3}{8}BC,$

$\dfrac{BN}{CN}=\dfrac{\frac{3}{8}BC}{\frac{5}{8}BC}=\dfrac{3}{5}$

4 0

3 years ago