All categories

3 years ago

6 + x is an example of a variable an expression a constant a formula

Mathematics

2 answers:

katrin [286]3 years ago

7 0

Answer:

an expression.

Step-by-step explanation:

if it was a variable it would be 6x but it not so it an expression

Sloan [31]3 years ago

3 0

Answer:varibale

Step-by-step explanation:

You might be interested in

NEED HELP FASTT

kotegsom [21]

Answer:

As the input increases, the output decreases for all values of x.

Step-by-step explanation:

The function is linear (a straight line) which heads down and to the right. There is no asymptote stopping the function from decreasing.

Formally, the behavior is described like so: as x approaches infinity, f(x) decreases

6 0

3 years ago

Find the area of the composite figure. Don't forget to use the word "square" and then the given units in your answer.

fgiga [73]

(11.5 x 34) + (16 x 16) = 256.5 meters squared :)

6 0

3 years ago

–12x + 9 = –15x solve for x

KonstantinChe [14]

Answer:

hi there!

the correct answer to question is: x=-3

Step-by-step explanation:

you move -12x to the other side by adding it on both sides to get: -3x=9 then you divide -3 on both sides to get x=-3

4 0

3 years ago

Read 2 more answers

Which graph shows a triangle and its reflection image across the x-axis?

Sav [38]

What are the options

8 0

3 years ago

Read 2 more answers

Determine if triangle ABC with coordinates A (0, 2), B (2, 5), and C (−1, 7) is an isosceles triangle. Use evidence to support y

kirill115 [55]

Answer:

The triangle ABC is an isosceles right triangle

Step-by-step explanation:

we have

The coordinates of triangle ABC are

A (0, 2), B (2, 5), and C (−1, 7)

we know that

An isosceles triangle has two equal sides and two equal internal angles

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the distance AB

substitute in the formula

$d=\sqrt{(5-2)^{2}+(2-0)^{2}}$

$d=\sqrt{(3)^{2}+(2)^{2}}$

$dAB=\sqrt{13}\ units$

step 2

Find the distance BC

substitute in the formula

$d=\sqrt{(7-5)^{2}+(-1-2)^{2}}$

$d=\sqrt{(2)^{2}+(-3)^{2}}$

$dBC=\sqrt{13}\ units$

step 3

Find the distance AC

substitute in the formula

$d=\sqrt{(7-2)^{2}+(-1-0)^{2}}$

$d=\sqrt{(5)^{2}+(-1)^{2}}$

$dAC=\sqrt{26}\ units$

step 4

Compare the length sides

$dAB=\sqrt{13}\ units$

$dBC=\sqrt{13}\ units$

$dAC=\sqrt{26}\ units$

$dAB=dBC$

therefore

Is an isosceles triangle

Applying the Pythagoras Theorem

$(AC)^{2} =(AB)^{2}+(BC)^{2}$

substitute

$(\sqrt{26})^{2} =(\sqrt{13})^{2}+(\sqrt{13})^{2}$

$26=13+13$

$26=26$ -----> is true

therefore

Is an isosceles right triangle

8 0

3 years ago