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VMariaS [17]
3 years ago
8

An inchworm ran into a log while on his way to the raspberry patch. The diameter of the log is 32 cm. How far did the inchworm t

ravel while on the log?

Mathematics
2 answers:
notsponge [240]3 years ago
7 0

Maybe I'm overthinking this one, but here's what I think it's talking about.
If I'm wrong, then I ought to at least get a few points for my talent at making
easy things difficult, and inventing obstacles to place in my own path.

-- The worm is 1 inch long.
-- The outside of the log is a cylinder.  Its cross-section is a
perfect circle with a circumference of 32-cm.
-- The axis (length) of the log is perpendicular (across) the path
that leads to raspberry nirvana.  
-- The ground is hard.  The log contacts the ground along a line,
and doesn't sink into it at all.

-- The worm sees the log ahead of him.  He continues crawling, until
he is directly under a point on the log that's 1-inch above him.
He then stands up to his full height, sticks his front legs to the log,
hoists himself up onto the bark, and starts to walk up and over it.

-- When he reaches a point on the other side of the log that's exactly 1-inch
above the ground, he hooks his sticky back feet to it, drops straight down to
the ground, and continues on his quest.

-- The question is:  What's the length of the part of the log's circumference
that he traveled between the two points that are exactly 1-inch off the ground ?

I thought I was going to be able to be able to talk through this, but I can't.
I need a picture.  Please see the attached picture.

Here comes the worm, heading from left to right.
He sees the log in front of him.
He doesn't bother going around it ... he knows he'll be able to get over it.

When he gets under the log, he starts standing straight up, trying to
grab onto the bark.  But he can't reach it.  He's too short, only 1 inch.

Finally, when he gets to point  'F', the bark is only 1" above him,
so he can hook on and haul himself up to point  'A'.

He continues on ... up, around, and over the log.

Eventually it dawns on him that the log won't last forever, and he'll
soon need to get down to the ground.  As he comes down the right
side of the log, he starts looking down.  It's too high.  He can't reach
the ground, and he's afraid to jump. 

Then he reaches point  'B'.  It's exactly 1-inch above the ground, and
he leaves the log and gets down.

What was the length of the path he followed on the log ... the long way,
over the top from  'A'  to  'B' ?

Here's what I did:

Draw radii from the center of the log to  'A'  and  'B' .
Each of them is 16 cm long (1/2 of the diameter).

Draw the radius from the center of the log to the ground (' E ').
It's 16 cm all the way.
Point  'D'  is 1 inch = 2.54 cm above the ground, so the
         vertical leg of each little right triangle is (16 - 2.54) = 13.46 cm.

There are two similar right triangles, back to back, inside the log.
They are  'CAD'  on the left, and  'CBD'  on the right.
I want to know the size of the angles at the top of each triangle.
(One will be enough, since they're equal angles.)

For each of those angles, the side adjacent to it is  13.46 cm.
And the hypotenuse of each right triangle is a radius, so it's 16 cm.
The cosine of those angles is  (adjacent/hypotenuse) = 13.46/16 = 0.84125 .
Each angle is  32.73 degrees.

Both of them put together add up to  65.45 degrees .

The full circumference of the log is  (pi)(D) = 32pi cm.
The short arc between 'A' and 'B' is  (65.45/360) of the full circumference.
The rest of the circumference is the distance that the worm crawled along it. 

     That's    (1 - 65.45/360) times (32 pi)  =  (0.818) x (32 pi) = <em>82.25 cm</em> .

Having already wasted enough time on this one in search of 5 points,
and then gone back through the whole thing to make corrections for
the customary worm crawling over the metric log, I'm not going to bother
looking for a way to check it.

That's my answer, and I'm sticking to it.

Cloud [144]3 years ago
5 0
Is the inchworm going around the log or over the log?

If it goes over the log, it simply travels the diameter, or 32 cm.

If it goes around the log, then it must travel half of the circumference, or 32*pi/2=16pi cm.

Note that this assumes that the inchworm and the raspberry patch are diametrically opposite.
You might be interested in
Rex paid $250 in taxes to the school district that he lives in this year. This year's taxes were a 12% increase from last year.
Verizon [17]

Answer:

$223.21

Step-by-step explanation:

Last year the amount of taxes was x, an unknown amount. That was 100% of last year's tax cost.

This year, taxes went up by 12%, so this year the taxes are 112% of what they were last year. This year the taxes were 112% of x, and they were $250.

112% of x = 250

1.12x = 250

x = 250/1.12

x = 223.21

Answer: $223.21

8 0
3 years ago
Please help with explanation and show work :)
ANEK [815]
3. A=bh/2
2A=bh
h=2A/b.

8. Dy-Cx=E
Dy=E+Cx
y=(E+Cx)/D
3 0
3 years ago
Could someone help me rnnn?
GuDViN [60]

Answer:

vertex = (0, -4)

equation of the parabola:  y=3x^2-4

Step-by-step explanation:

Given:

  • y-intercept of parabola: -4
  • parabola passes through points: (-2, 8) and (1, -1)

Vertex form of a parabola:  y=a(x-h)^2+k

(where (h, k) is the vertex and a is some constant)

Substitute point (0, -4) into the equation:

\begin{aligned}\textsf{At}\:(0,-4) \implies a(0-h)^2+k &=-4\\ah^2+k &=-4\end{aligned}

Substitute point (-2, 8) and ah^2+k=-4 into the equation:

\begin{aligned}\textsf{At}\:(-2,8) \implies a(-2-h)^2+k &=8\\a(4+4h+h^2)+k &=8\\4a+4ah+ah^2+k &=8\\\implies 4a+4ah-4&=8\\4a(1+h)&=12\\a(1+h)&=3\end{aligned}

Substitute point (1, -1) and ah^2+k=-4 into the equation:

\begin{aligned}\textsf{At}\:(1.-1) \implies a(1-h)^2+k &=-1\\a(1-2h+h^2)+k &=-1\\a-2ah+ah^2+k &=-1\\\implies a-2ah-4&=-1\\a(1-2h)&=3\end{aligned}

Equate to find h:

\begin{aligned}\implies a(1+h) &=a(1-2h)\\1+h &=1-2h\\3h &=0\\h &=0\end{aligned}

Substitute found value of h into one of the equations to find a:

\begin{aligned}\implies a(1+0) &=3\\a &=3\end{aligned}

Substitute found values of h and a to find k:

\begin{aligned}\implies ah^2+k&=-4\\(3)(0)^2+k &=-4\\k &=-4\end{aligned}

Therefore, the equation of the parabola in vertex form is:

\implies y=3(x-0)^2-4=3x^2-4

So the vertex of the parabola is (0, -4)

5 0
2 years ago
Read 2 more answers
[WILL MARK BRAINLIEST]
notka56 [123]

Answer:

The answer is 4 cm

Step-by-step explanation:

The answer is 4 cm because the formula for a cone is V=π r^2 h/3, and the formula for a cylinder is V=π r^2 h, and the cone's volume is V≈301.59, and, a cylinder with

r=4 and h=6 is V≈301.59

Therefore the answer is

4 cm

I hope this has satisfied you.

8 0
3 years ago
Read 2 more answers
43. Can you place ten lumps of sugar in three empty cups so that
erma4kov [3.2K]

Answer: No.

Step-by-step explanation:

We have 10 lumps of sugar, and we want to divide them into 3 cups, in such a way that there is an odd number of lumps in each cup.

This only can happen if we have 3 odd numbers such that the addition is equal to 10.

Now 10 is an even number, remember that even numbers can be written as:

2*k

where k is an integer number.

And odd numbers can be written as:

2*n + 1

where n is an integer.

Then we have 3 odd numbers, let's call them:

(2*n + 1), (2*k + 1) and (2*p + 1).

Now let's add them:

(2*n + 1) + (2*k + 1) + (2*p + 1).

2*(n + k + p) + 1 + 1+ 1

2*(n + k + p) + 2 + 1 =

2*(n + k + p + 1) + 1.

Now, the number n + k + p + 1 is an integer number, let's call it X, then we have that the addition of the 3 odd numbers is:

2*X + 1

This is an odd number

So for any 3 odd numbers that we add together, the result will always be an odd number.

Then is impossible to add 3 odd numbers and get 10 as the result (Again, 10 is an even number).

Then is not possible to have an odd number of lumps in each cup.

5 0
3 years ago
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