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3 years ago

Which sets of values belong to the domain of range of a relation?

Mathematics

2 answers:

aliina [53]3 years ago

5 0

Domain is all X values, and range is the Y values.

Goryan [66]3 years ago

4 0

For Domain:

values for the independent variable

input values

For range:

values for the dependent variable

output values

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Solve the system of equations. If the system does not have one unique solution, determine the number of solutions to the system.

Licemer1 [7]

See photo. Hope it helps

3 0

3 years ago

When will these two lines intersect <br>C. y=-2x +3<br>y = x

Andrej [43]

Answer:

The intersection point of the given lines is (1,1).

Step-by-step explanation:

Here, the given equations are:

y = - 2 x + 3

y = x

Substitute y = x in the first equation,

y = -2 x + 3 becomes y = -2 (y) + 3

or, y + 2 y = 3

or, 3 y = 3 ⇒ y = 3/3 = 1

⇒ y = 1

Now, as x = y ⇒ x = 1

Hence, the intersection points of the given lines is (1,1).

8 0

3 years ago

How many, Inch cubes fit into a rectangular prism that measures 4 Inches long, 5 Inches wide, and

kramer

780 (or c)

I think this is correct because 4 x 5 = 20 x 5 = 100 and 100 x 5 = 500

500 + 280 = ????

780

8 0

2 years ago

A window shaped like a parallelogram has an area of 18 1/3 square feet. The height of the window is 3 1/3 feet. How long us the

Lyrx [107]

The area of a parallelogram can be found in the same way as that of a rectangle, so base x height. To find the base we have to do area ÷ height or 18 1/3 ÷ 3 1/3= 5.5 feet

4 0

3 years ago

Theorem: A line parallel to one side of a triangle divides the other two proportionately.

Kisachek [45]

Answer:

Segment BF = 16

Step-by-step explanation:

The given theorem states that a line parallel to one side of a triangle divides the other two sides proportionately

The given theorem is the Triangle Proportionality Theorem

According to the theorem, given that segment DE is parallel to segment BC, we have;

$\dfrac{AD}{BD} = \dfrac{AE}{EC}$

Therefore;

$BD = \dfrac{AD}{\left(\dfrac{AE}{EC} \right) } = AD \times \dfrac{EC}{AE}$

Which gives;

$BD = 6 \times \dfrac{18}{12}= 9$

Similarly, given that EF is parallel to AB, we get;

$\dfrac{AE}{EC} = \dfrac{BF}{FC}$

Therefore;

$BF = FC \times \dfrac{AE}{EC}$

Which gives;

$BF = 24 \times \dfrac{12}{18} = 16$

Therefore, the statement that can be proved using the given theorem is segment BF = 16.

8 0

3 years ago