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 travel while on the log?
Answers
Answer 1
Answer:
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.
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.
Answer 2
Answer:
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) = 82.25 cm .
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.
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) = 82.25 cm .
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.
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Answers
Answer:
.................................
Step-by-step explanation:
Answers
Answer:
Step-by-step explanation:
Alright, lets get started.
The side length of square is given as 2.5.
So, the perimeter if square will be :
Once the side is doubled, means new dimension of side of square will be :
So the new perimeter will be :
The original perimeter is 10.
The new perimeter is 20.
It means the perimeter is twice the original perimeter. : Answer D
Hope it will help :)
T=4.5
Simplifying3t + 4 = t + 13
Reorder the terms:4 + 3t = t + 13
Reorder the terms:4 + 3t = 13 + t
Solving4 + 3t = 13 + t
Solving for variable 't'.
Move all terms containing t to the left, all other terms to the right.
Add '-1t' to each side of the equation.4 + 3t + -1t = 13 + t + -1t
Combine like terms: 3t + -1t = 2t4 + 2t = 13 + t + -1t
Combine like terms: t + -1t = 04 + 2t = 13 + 04 + 2t = 13
Add '-4' to each side of the equation.4 + -4 + 2t = 13 + -4
Combine like terms: 4 + -4 = 00 + 2t = 13 + -42t = 13 + -4
Combine like terms: 13 + -4 = 92t = 9
Divide each side by '2'.t = 4.5
Simplifyingt = 4.5
I got help from a website called geteasysolution
Simplifying3t + 4 = t + 13
Reorder the terms:4 + 3t = t + 13
Reorder the terms:4 + 3t = 13 + t
Solving4 + 3t = 13 + t
Solving for variable 't'.
Move all terms containing t to the left, all other terms to the right.
Add '-1t' to each side of the equation.4 + 3t + -1t = 13 + t + -1t
Combine like terms: 3t + -1t = 2t4 + 2t = 13 + t + -1t
Combine like terms: t + -1t = 04 + 2t = 13 + 04 + 2t = 13
Add '-4' to each side of the equation.4 + -4 + 2t = 13 + -4
Combine like terms: 4 + -4 = 00 + 2t = 13 + -42t = 13 + -4
Combine like terms: 13 + -4 = 92t = 9
Divide each side by '2'.t = 4.5
Simplifyingt = 4.5
I got help from a website called geteasysolution
Answers
Answer:
Average speed(S) is given by:
....[1]
where,
D is the total distance and t is the time.
As per the statement:
A gazelle that runs 140 meters in 5.0 seconds.
⇒D = 140 meters and t = 5 seconds
Substitute in [1] we have;
Simplify:
⇒S = 28 m/sec
Therefore, average speed of a gazelle is, 28 m/sec
You would multiply to find the average meters/minute.
140 x 5.0 = 700.
But, if you're looking for average m/s you would divide.
140/5 = 28.
If you want meters/hour, you have to multiply the average meters/minute by 60.
140 x 5 x 60 = 4,200.
I'm not sure which one you're looking for but I hope something I wrote helps!
140 x 5.0 = 700.
But, if you're looking for average m/s you would divide.
140/5 = 28.
If you want meters/hour, you have to multiply the average meters/minute by 60.
140 x 5 x 60 = 4,200.
I'm not sure which one you're looking for but I hope something I wrote helps!
Answers
Well, first of all, every positive number has two square roots.
Their digits are the same, but one is positive and one is negative.
The square root of 343 is an irrational number, so it can't ever be
exactly written with digits. As a decimal, it never ends or repeats,
and as a fraction, it can't be written at all.
The positive square root starts out 18.52025... and keeps going forever.
The other square root is the negative of that same number.
When you multiply either of them by itself, the product is close
to 343 . The more decimal places you include, the closer it gets,
but it can never be exactly 343 .
Answers
The above question cannot be solved because the sum of five odd numbers is an odd number
No answer to the above question because the sum of five odd number is a odd number. Here 30 is not a odd number.
Answers
estimate- 4,520
if you multiply -371954
Adding- 1779
if you multiply -371954
Adding- 1779
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