1/5, 2/3, 5/8 from least to greatest on a number line

A man starts at point zero.He walks forward four paces to point 4. He returns to point zero and continues walking back four more

Svet_ta [14]

Answer:

New point will be (-4) or (-4, 0).

Step-by-step explanation:

"A man starts at a point zero".

If we locate this point on a graph we call it origin (0, 0).

"He walks forward four paces to point 4".

Then he reaches the point (4, 0) which is 4 units apart from the origin.

"He returns to the point zero".

That means man comes back to the origin (0, 0) by covering 4 units back.

"He continues walking back for 4 more paces"

That means man doesn't stop at zero and continues walking for 4 paces in the same direction.

On the graph he moves towards the negative direction of the x-axis.

So on graph the new point will be (-4, 0).

In more simpler way he will be at (-4) point.

5 0

3 years ago

A builder was building a fence. In the morning, he worked for 25 of an hour. In the afternoon, he worked for 910 of an hour. How

Furkat [3]

9514 1404 393

Answer:

(9/10)/(2/5) = ratio of work times

2 1/4 times as long

Step-by-step explanation:

Given:

2/5 hours work in the morning

9/10 hours work in the afternoon

Find:

an equation for how many times as long was afternoon work compared to morning work

the equation solution

Solution:

a/m = r . . . . ratio of afternoon work to morning work

(9/10)/(2/5) = r . . . equation

(9/10)/(4/10) = r

9/4 = r = 2 1/4 . . . solution

The builder worked 2 1/4 times as long in the afternoon as in the morning.

8 0

3 years ago

A Suppose 13-year and 17-year cicadas both appear this year. When will both

meriva

Answer:Stragglers can emerge 1 or 4 years early or 1 or 4 years late. Don't be surprised if you see some periodical cicadas emerge earlier than planned this year. 17-year brood members are most likely to straggle 4 years early, and 13-year brood members are most likely to straggle 4 years late.

Step-by-step explanation:

4 0

3 years ago

How many solutions does 4(x-1)-x = 3(x+5)-11 have

bonufazy [111]

Answer:

None.

Step-by-step explanation:

4(x-1)-x = 3(x+5)-11

4x-4-x=3x+15-11

we combine like terms.

4x-x-4=3x+14

3x-4=3x+14

we then isolate the variables.

3x-4=3x+14

+4 +4

3x=3x+18

-3x -3x

0≠18

No solutions

6 0

3 years ago

Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. (Enter your answers using inte

Mashcka [7]

Answer:

f'(x) > 0 on $(-\infty ,\frac{1}{e})$ and f'(x)<0 on $(\frac{1}{e},\infty)$

Step-by-step explanation:

1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:

$f'(x)>0$

To find its decreasing interval :

$f'(x)$

2) Then let's find the critical point of this function:

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0$

2.2 Solving for x this equation, this will lead us to one critical point since x' is not defined for Real set, and x'' $=\frac{1}{e}$ ≈0.37 for e≈2.72

$x^{2x}=0 \ni \mathbb{R}\\(ln(x)+1)=0\Rightarrow ln(x)=-1\Rightarrow x=e^{-1}\Rightarrow x\approx 2.72^{-1}x\approx 0.37$

3) Finally, check it out the critical point, i.e. f'(x) >0 and below f'(x)<0.

8 0

3 years ago