All categories

4 years ago

The drop down options are fell or rose PLEASE HELP 6th grade math btw

Mathematics

2 answers:

Yanka [14]4 years ago

7 0

fell look its a negative so the amount dropped try that

PIT_PIT [208]4 years ago

3 0

The answer is fell

Hope it is correct and helps

You might be interested in

Which graph represents (,y)-pairs that make the equation y = 2x true?

Veseljchak [2.6K]

Answer:

Pairs (2,4), (4,8) and (5,10) makes the equation y=2x true

Step-by-step explanation:

By using the equation y=2x, eval the following points (gathered from the attached picture)

1. (2,4) which leaves us 4=2*2

2. (4,8) which leaves us 8=4*2

3. (5,10) which leaves us 10=5*2

All points from this graph satisfies the equation y=2x

8 0

3 years ago

a bag contains 7 red marbles, 9 white marbles, and 5 blue marbles. randomly choose two marbles, one at a time, and without repla

Natalija [7]

Answer: could you please be more clear..?

Step-by-step explanation:

5 0

3 years ago

Multiply.

wolverine [178]

So you're initial equation is $(2/6)*24=m$ . Well since $2/6=1/3$ , plug in 24 for the 1 in $1/3$ . So $24/3=m$ , and $24/3=8$ so $m=8$ .

7 0

4 years ago

Consider the line 5x-8y=8 what is the slope of a line perpendicular to this line what is the slope of a line parallel to this li

Sophie [7]

$\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2,\ \text{then}\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\\\\text{Convert the equation to the slope-intercept form}\ y=mx+b.\\\\5x-8y=8\qquad\text{subtract}\ 5x\ \text{from both sides}\\\\-8y=-5x+8\qquad\text{divide both sides by (-8)}\\\\y=\dfrac{5}{8}x-1\to m_1=\dfrac{5}{8}\\\\perpendicular:\ m_2=-\dfrac{1}{\frac{5}{8}}=-\dfrac{8}{5}\\\\parallel:\ m_2=\dfrac{5}{8}$

8 0

3 years ago

Find the area of each regular polygon. Round your answer to the nearest tenth if necessary.

tatuchka [14]

*I am assuming that the hexagons in all questions are regular and the triangle in (24) is equilateral*

(21)

Area of a Regular Hexagon: $\frac{3\sqrt{3}}{2}(side)^{2} = \frac{3\sqrt{3}}{2}*(\frac{20\sqrt{3} }{3} )^{2} =200\sqrt{3}$ square units

(22)

Similar to (21)

Area = $216\sqrt{3}$ square units

(23)

For this case, we will have to consider the relation between the side and inradius of the hexagon. Since, a hexagon is basically a combination of six equilateral triangles, the inradius of the hexagon is basically the altitude of one of the six equilateral triangles. The relation between altitude of an equilateral triangle and its side is given by:

$altitude=\frac{\sqrt{3}}{2}*side$