MATHEMATICS HIGH SCHOOL

Please help me to answer this question thank you otherwise I my teacher will be so mad .
Please help me to answer this question thank you otherwise - 1

Answers

Answer 1
Answer:

Answer:

2/3 ≈ 0.666667

5/11 ≈ 0.454545

5/6 ≈ 0.833333

6/7 ≈ 0.857143

Step-by-step explanation:

Easiest and fastest way to do this is to use a calculator to find your decimals and then approximate.
Answer 2
Answer:

Answer:

2/3 = 2 divided by 3 so 2 divided by 3 = 0.666667. We round up because it is a repeating decimal which goes on forever

5/11= 5 divided by 11 = 0.454545

5/6 = 5 divided by 6 =0.833333

6/7 = 6 divided by 7 = 0.857143

rounded up because it is an infinite decimal

Hope this helps

Step-by-step explanation:

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How do you write 35 over 20 As a percentage

1,352 pounds is what fraction short of a ton? A) 169/250             B) 77/100               C)13/20             D140/223

Answers

A \ short \ ton \ is \ 2 \ 000 pounds \n \n(1 \ 352)/(2 \ 000) = (169)/(250) \ \ \ | divided \ numerator \ and \ denominator \ by \ 8 \n \n \n Answer : \ A) \ \ ( 169)/(250 )
1352/2000 so then you simplify by dividing both sides to get
A) 169/250  

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

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.

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.

Two similar pyramids have corresponding dimensions in the ratio 3:5 Find the ratio of their volumes

Answers

Answer:27:125

Step-by-step explanation:

Given

Ratio of corresponding dimensions is in the ratio of 3:5

If the two pyramids are similar then the ratio of their volumes is proportional to the cube of their side ratio

i.e.

\Rightarrow (V_1)/(V_2)=((l_1)/(l_2))^3

\Rightarrow (V_1)/(V_2)=((3)/(5))^3

\Rightarrow (V_1)/(V_2)=(27)/(125)

Thus the ratio of their volumes is 27:125

Solve for x and express your answer as a logarithm.2(10^(3x) ) = 24


Simplify your answer as much as possible.


Use the following notations where necessary:

• For fractions, use the / symbol to separate numerator and denominator, like this: 42/53

• For logs with a base of 10, such as log107, just write the log without the base and place the value in parentheses, like this: log(7)

• For logs with a base other than 10, write the base after an underscore then place the value in parentheses. For example, to write log27, write it like this: log_2(7)

Answers

2\cdot10^(3x)=24\ \ \ \ |divide\ both\ sides\ by\ 2\n\n10^(3x)=12\iff\log(10^(3x))=\log(12)\n\n3x\log(10)=\log(12)\n\n3x=\log(12)\ \ \ \ |divide\ both\ sides\ by\ 3\n\n\boxed{x=(\log(12))/(3)}\to(\log(12))/(3)=(1)/(3)\log(12)=\log\left(12^(1)/(3)\right)=\boxed{\log\sqrt[3]{12}}

Use:\n\log_ab^c=c\log_ab\n\log_aa=1\na^(1)/(n)=\sqrt[n]{a}\n-------------------\n\nAnswer:\boxed{x=(\log(12))/(3)\ other\ form\ x=\log(\sqrt[3]{12})}

Which graphs represent functions?

Answers

Answer:

A. Only graph B and D

Step-by-step explanation:

Hi, to answer this question we have to analyze the options given:

A function has only one output value (y) for each input value.(x)

In other words, If we draw a vertical line (anywhere on the graph) that intersects the graph in two points or more, then the graph does not represent a function because that x value has more than one output(y).

So, the correct option is:

A. Only graph B and D

What is the solution of log3x − 12 729 = 6?x = −5
x = −3
x = 3
x = 5

*"3x- 12" is next to log, but it's in the bottom, like an exponent, but instead of it being on top, it's below.

Answers

Domain:\nD:3x-12 > 0\ and\ 3x-12\neq1\ \ \ |add\ 12\ to\ both\ sides\n3x > 12\ and\ 3x\neq13\ \ \ \ |divide\ both\sides\ by\ 3\nx > 4\ and\ x\neq(13)/(3)\n\boxed{x\in\left(4;\ (13)/(3)\right)\ \cup\ \left((13)/(3);\ \infty\right)}\n----------------------\n\n\log_(3x-12)729=6\n--------\n\log_ab=c\iff a^c=b\n----------\n\log_(3x-12)729=6\iff(3x-12)^6=729\n\n(3x-12)^6=3^6\iff3x-12=3\ \ \ |add\ 12\ to\ both\ sides\n3x=15\ \ \ \ |divide\ both\ sides\ by\ 3\n\boxed{\boxed{x=5\in D}}
log _(3x-12) ^(729) = 6
(3x-12)^(6) = 7293x - 12 = \sqrt[6]{729} = 33x= 15x=5
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