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3 years ago

2 more i will give Brainliest

Mathematics

2 answers:

bezimeni [28]3 years ago

4 0

1 and 1/2 is the scale factor

AveGali [126]3 years ago

4 0

Yo pls I need it it to answer a. Quads

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2x-6y=12 and -5x+15y=-30 elimination

Reptile [31]

Answer:

unsolvable because the second equation is just -2.5 times the first equation.

Step-by-step explanation:

2x-6y=12 then 2x=12+6y x=6+3y

-5x+15y=-30

substitute x

-5(6+3y)+15y=-30

-30-15y+15y=-30

y=y

8 0

3 years ago

noah is baking a two layer cake in which the bottom layer is a circle and the top layer is a triangle if line ab

padilas [110]

Since , segment AB ≅ segment BC as their arcs are congruent ; therefore , ΔABC is an isosceles triangle .

Properties of a circle :

If a circle is circumscribed around a triangle, then the vertices of the

1. Triangle divides the circle into three arcs

2. The three sides of the triangle are the chords of the circle

3. Each chord in the circle subtended by an arc

4. If two arcs equal, then they are subtended equal chord

∵ Circumscribed circle D around triangle ABC

- The vertices of triangle ABC lie on the circumference of the circle

AB , BC , AC are the chords of the circle

Arcs AB and AC are congruent

Arc AB ≅ arc AC

Arc AB subtended the chord AB

Arc AC subtended the chord AC

AB = AC

AB = 6 inches

AC = 6 inches

- The triangle with two equal sides is called an isosceles triangle

∴ Δ ABC is an isosceles triangle .

Noah know that the top layer of his cake is an isosceles triangle

* segment AB ≅ segment BC because their arcs are congruent;

Therefore, ΔABC is an isosceles triangle.

To learn more on circle properties :

brainly.com/question/4244936

#SPJ4

8 0

1 year ago

Which of the following tables does not represent a linear relationship between x and y?

vovangra [49]

Answer:

The first one is not liner

Step-by-step explanation:

This isn't linear because the y values increase inequally. It adds 2 and then 3 instead of keeping a constant rate

6 0

3 years ago

Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2

Mkey [24]

For this case, we have to:

By definition, we know:

The domain of $f (x) = \sqrt [3] {x}$ is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root. Thus, it will always be defined.

So, we have:

$y = \sqrt [3] {x-2}$ with $x = 0$ : $y = \sqrt [3] {- 2}$ is defined.

$y = \sqrt [3] {x+2}$ with $x = 0:\ y = \sqrt [3] {2}$ is also defined.

$f (x) = \sqrt {x}$ has a domain from 0 to ∞.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

$y = \sqrt {x-2}$ is not defined, the term inside the root is negative when $x = 0$ .