If it takes 3 days for 10 workers to finish building one house, how many days will it take 15 workers to finish four houses?
Answers
so
-- It takes 10 workers 1 day to build 1/3 of a house.
so
-- It takes 1 worker 1 day to build 1/30 of a house.
You got 15 workers and you need 4 houses ?
Well, as we just calculated . . .
-- It takes 1 worker 1 day to build 1/30 of a house.
so
-- It takes 15 workers 1 day to build 15/30 = 1/2 of a house.
so
-- It takes 15 workers 2 days to build 1 whole house.
so
-- It takes 15 workers 8 days to build 4 houses.
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Answers
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.
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.
Answers
5x - 5x = 6x² - 5x - 3
0 = 6x² - 5x - 3
x = -(-5) ± √((-5)² - 4(6)(-3))
2(6)
x = 5 ± √(25 + 72)
12
x = 5 ± √(97)
12
x = 5 + √(97) U x = 5 - √(97)
12 12
The value of x are; x = 5 + √(97)/12 and x = 5 - √(97)/12.
What is a quadratic equation?
A quadratic equation is the second-order degree algebraic expression in a variable.
The standard form of this expression is ax² + bx + c = 0 where a. b are coefficients and x is the variable and c is a constant.
Given;
5x = 6x² - 3
Subtract 5x on both sides
5x - 5x = 6x² - 5x - 3
0 = 6x² - 5x - 3
x = -(-5) ± √((-5)² - 4(6)(-3)) / 2(6)
x = 5 ± √(25 + 72)/ 12
x = 5 ± √(97)/ 12
The value of x are;
x = 5 + √(97)/12
x = 5 - √(97)/12
Learn more about quadratic equations;
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Answers
Answers
Answer:
$26.25 is the cost of 3.5 square meters canvas.
Step-by-step explanation:
Cost of one unit square canvas is given as $7.50 and we have to calculate the cost of 3.5 square meter cost of the canvas.
We can use the unitary method to solve this question.
∵ Cost of 1 square meter canvas = $7.50
∴ Cost of 3.5 square meter canvas = 7.5 × 3.5
= $26.25
Therefore, cost of 3.5 square meter of the canvas is $26.25.
Answers
A 36
B 8
C 7
D 13