MATHEMATICS HIGH SCHOOL

Which value is the best estimate for
√
94

Answers

Answer 1
Answer:

Answer:

9.7

Step-by-step explanation:
Answer 2
Answer: V95=9.7 is the answer

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Which statements are true regarding the area of circle D? Select two options. The area of the circle depends on the square of the radius.
The area of circle D is 36Pi centimeters squared.
The area of circle D is 324Pi centimeters squared
The area of the circle depends on the square of pi.
The area of the circle depends on the central angle.

Answers

Answer:

The ratio of the measure of central angle PQR to the measure of the entire circle is 1/8

The area of the shaded sector depends on the length of the radius.

The area of the shaded sector depends on the area of the circle.

Step-by-step explanation:

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.

What polynomial identity should be used to prove that 16^2=(10+6)^2? (A. Difference of Cubes; B. Difference of Squares; C. Square of Binomial; D. Sum of Cubes)

Answers

Answer:

C. Square of Binomial

Step-by-step explanation:

To prove the identity you should use square of Binomial that states the following:

(a+b)^(2)=a^(2)+2ab+b^(2)

Lets prove it, so first take the equation to solve:

(10+6)^(2)

Then square the first term:

(10+6)^(2)=10^(2)

Then multiply by 2 the first and second terms:

(10+6)^(2)=10^(2)+2(10)(6)

Finally square the second term:

(10+6)^(2)=10^(2)+2(10)(6)+6^(2)

Solve the values:

(10+6)^(2)=100+2(10)(6)+6^(2)

(10+6)^(2)=100+120+6^(2)

(10+6)^(2)=100+120+36

(10+6)^(2)=256

And prove the polynomial identity:

16^(2)=256
not difference of cubes since it is 2nd degree
not difference of squares since it is plus
not sum of cubes because 2nd degreee

answer is square of binomial (why do we even need this property, oh well)


C

Insert < ​,>​, or = between the pair of numbers to form a true statement. 0.32 and 0.35

Answers

Answer:   0.32 < 0.35

Think of it like saying 32 < 35. Then stick a decimal point in front of each number.

Please help me this is due tonight

Answers

Answer:

  B = (2, -5)

Step-by-step explanation:

The problem requirements mean ...

  B - A = (2/3)(C - A)

  B = A + (2/3)C - A

  B = (1/3)A + (2/3)C = (A +2C)/3

  B = ((-6, 1) +2(6, -8))/3 = (-6+12, 1-16)/3 = (6, -15)/3

  B = (2, -5)

A construction worker tosses a brick from a tall building. The brick's height (in meters above the ground) t tt seconds after being thrown is modeled by h ( t ) = − 5 t 2 + 20 t + 105 h(t)=−5t 2 +20t+105h, left parenthesis, t, right parenthesis, equals, minus, 5, t, squared, plus, 20, t, plus, 105 Suppose we want to know the height of the brick above the ground at its highest point. 1) Rewrite the function in a different form (factored or vertex) where the answer appears as a number in the equation. h ( t ) = h(t)=h, left parenthesis, t, right parenthesis, equals 2) At its highest point, how far above the ground was the brick?

Answers

The vertex form of the equation is h(t) = -5 ( t - 2)² + 104.

At its highest point, the brick was 101 meters above the ground.

What is the vertex form of a quadratic function?

The factored form of a quadratic function is given as:

f(x) = a(x – h)2 + k

where a, h, and k are constants.

We have,

h(t) = -5t² + 20t + 105

h(t) = -5 ( t² - 4t ) + 105

h(t) = -5 ( t² - 4t + 2² ) - 2² + 105

h(t) = -5 ( t - 2 )² - 4  +  105

h(t) = -5 ( t - 2 )² +  101 _____(1)

The vertex form of the equation is h(t) = -5 ( t - 2)² + 104.

From equation (1) we can say that at its highest point the brick was at

101 meters above the ground.

Thus,

The vertex form of the equation is h(t) = -5 ( t - 2)² + 104.

At its highest point, the brick was 101 meters above the ground.

Learn more about quadraticequations here:

brainly.com/question/18804178

#SPJ5

Answer:

h(t)=−5(t−2)^2+125, 125

Step-by-step explanation:
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