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In a standard dictionary, where can you find the key to pronunciation marks? A. At the bottom of each page B. Within the definitions C. In the front of the dictionary D. In an appendix

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# Tag: bottom

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The western part of South America is a region that is geologically very active. This is due to the boundaries between the South American tectonic plate with the Nazca tectonic plate, and also with the Pacific tectonic plate. Since both, the Nazca and the Pacific plates are oceanic plates, they are heavier, and they subduct below the lighter plate of South America, thus enabling the pressure and temperature from the mantle to come out towards the surface on this boundary lines. This creates lots of earthquakes, as well as volcanic activity.

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To solve this problem,lets say that

X = the weight of the machine components.

X is normally distributed with mean=8.5 and sd=0.09

We need to find x1 and x2 such that

P(X<x1)=0.03 and P(X>x2)=0.03

Standardizing:

P( Z< (x1 – 8.5)/0.09 ) =0.03

P(Z > (x2 – 8.5)/0.09 ) =0.03.

From the Z standard table, we can see that approximately P

= 0.03 is achieved when Z equals to:

z = -1.88 and z= 1.88

Therefore,

P(Z<-1.88)=0.03 and P(Z>1.88)=0.03

So,

(x1 – 8.5)/0.09 = -1.88 and

(x2 – 8.5)/0.09 =1.88

Solving for x1 and x2:

x1=-1.88(0.09) + 8.5 and

x2=1.88(0.09) + 8.5

Which yields:

**x1 = 8.33 g**

**x2 = 8.67 g**

Answer: The bottom 3 is separated by the weight

8.33 g and the top 3 by the weight 8.67 g.

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Which equation can be used to calculate the area of the shaded triangle in the figure below? A rectangle is shown with a width of 6 ft and a length of 16 ft. There is a diagonal line through the rectangle, and the bottom half is shaded in grey. choices 2(16 × 6) = 192 square feet 2(16 + 6) = 44 square feet 1 over 2.(16 × 6) = 48 square feet 1 over 2.(6 + 16) = 11 square feet

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A pair of parallel lines is cut by a transversal: A pair of parallel lines cut by a transversal is shown. The inside bottom right angle is labeled 80 degrees. The alternate interior angle is cut by a ray and is labeled as x on one side and 20 degrees on the other. What is the measure of angle x?

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According to this diagram what is tan 67 It’s a right triangle facing left. hypotenuse: 13 opposite:5 adjacent: 12 The angle to the far left is 23 degrees, the one to the far right at the bottom is 90, and the one on the top is 67. options: 12, 5/13, 12/13, 5/12, 13/5, 13/12 Please explain. Thanks so much

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Eric drove his fishing boat into a large cove and dropped the anchor in 52 feet of water the anchor descends at a constant rate of 3.8 feet per second, how long will it take for the anchor to hit the bottom of the lake? Round your answer to the nearest tenth If of a second if necessary.

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Hey there!

**Correct answer is D. Comma or Period Inside Rule**

A. Question mark or exclamation point inside: those are not really necessary.

B. Colon or semicolon: not, a semicolon would divide the whole sentence and it would be shorten; a colon would work but after requested, when there is already a comma.

C. Question mark or Exclamation Point Outside Rule: would not work, becase it is an very polite and affirmative sentence.

**D: A comma or period inside rule: actually, just a comma would work. Please, Cooper’s dad requested, go… Cooper’s dad requested must be in between commas as it is a vocative.**

Hope this helps

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**Answer:**

Option C – BD=76 cm

**Step-by-step explanation:**

**Given :** You are designing a diamond-shaped kite. you know that AD = 44.8 cm, DC = 72 cm, and AC = 84.8 cm.

**To find : **How long BD should it be?

**Solution : **

First we draw a rough diagram.

The given sides were AD = 44.8 cm, DC = 72 cm and AC = 84.8 cm.

According to properties of kite

**Two disjoint pairs of consecutive sides are congruent.**

**So,** AD=AB=44.8 cm

DC=BC=72 cm

**The diagonals are perpendicular.**

**S**o, AC ⊥ BD

Let O be the point where diagonal intersect let let the partition be x and y.

AC= AO+OC

AC= …….[1]

Perpendicular bisect the diagonal BD into equal parts let it be z.

BD=BO+OD

BD=z+z

**Applying Pythagorean theorem in ΔAOD**

where H=AD=44.8 ,P= AO=x , B=OD=z

………[2]

**Applying Pythagorean theorem in ΔCOD**

where H=DC=72 ,P= OC=y , B=OD=z

…………[3]

**Subtract [2] and [3]**

……….[4]

**Add equation [1] and [4], to get values of x and y**

**Substitute x in [1]**

**Substitute value of x in equation [2], to get z**

**We know, **BD=z+z

BD= 38.06+38.06

BD= 76.12

**Nearest to whole number **BD=76 cm

**Therefore, Option c – BD=76 cm is correct.**

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**Answer:**

Dogs can be strong companions, but they also have needs that many people forget to address.

**Explanation:**

The author’s claim is that many people forget to address the needs of their dogs. The author continues to talk about some of those needs, like regular washing and grooming to avoid becoming matted and sick. The author is not simply talking about washing and grooming, but other needs the dogs might have as well. These could include things like walking, a proper diet, or brushing their teeth.

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**Answer: The correct option is (C). 10 = square root of the quantity of x minus 8 all squared plus y minus 9 all squared **

**Step-by-step explanation: **Given that the segment AB has point A located at (8, 9). The distance from A to B is 10 units.

We are to **select the correct option that could be used to calculate the coordinates for point B.**

**Let, (x, y) be the co-ordinates of point B.**

**According to distance formula, the distance between two points (a, b) and (c, d) is given by**

**Therefore, the distance between the points A(8, 9) and B(x, y) is given by**

**Since, distance between A and B is 10 units, so**

**d = 10.**

**Therefore,**

**Thus, the correct statement is**

** 10 = square root of the quantity of x minus 8 all squared plus y minus 9 all squared.**

**Option (C) is correct. **

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**Answer:**

a) The time the police officer required to reach the motorist was 15 s.

b) The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) The total distance traveled by the officer was 225 m.

**Explanation:**

The equations for the position and velocity of an object moving in a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

a)When the officer reaches the motorist, the position of the motorist is the same as the position of the officer:

x motorist = x officer

Using the equation for the position:

x motirist = x0 + v · t (since a = 0).

x officer = x0 + v0 · t + 1/2 · a · t²

Let´s place our frame of reference at the point where the officer starts following the motorist so that x0 = 0 for both:

x motorist = x officer

x0 + v · t = x0 + v0 · t + 1/2 · a · t² (the officer starts form rest, then, v0 = 0)

v · t = 1/2 · a · t²

Solving for t:

2 v/a = t

t = 2 · 15.0 m/s/ 2.00 m/s² = 15 s

The time the police officer required to reach the motorist was 15 s.

b) Now, we can calculate the speed of the officer using the time calculated in a) and the equation for velocity:

v = v0 + a · t

v = 0 m/s + 2.00 m/s² · 15 s

v = 30 m/s

The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) Using the equation for the position, we can find the traveled distance in 15 s:

x = x0 + v0 · t + 1/2 · a · t²

x = 1/2 · 2.00 m/s² · (15s)² = 225 m

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A ladder is leaning against a wall. The top of the ladder is 9feet above the ground. If the bottom of the ladder is moved 3ft farther from the wall, the ladder will be lying flat on the ground, still touching the wall. How long, in feet, is the ladder?

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distance between (x1,y1) and (x2,y2) is

so distance between (-2,4) and (18,-6) is

D=10√5

so in ratio 3:7

3+7=10

3/10 of 10√5 is 3√5

I’m thinkin you want R then Q then S such that RQ:QS=3:7

so

distance from R to Q is 3√5

Q is (x,y)

R is (-2,4)

D=3√5

square both sides

now, ithas to be on the line that R and S are on

do some simple math

slope is rise/run=-10/20=-1/2

y=-1/2x+b

4=-1/2(-2)+b

4=1+b

3=b

y=-1/2x+b

sub that for y in our other equation ()

I’m too lazy to show you expanstion and whatnot so I’ll give you the solution

we get (after some manipulation)

0=x²+4x-32

what 2numbers multiply to get -32 and add to get 4?

-4 and 8

0=(x-4)(x+8)

set to zero

0=x-4

4=x

0=x+8

-8=x

but wait, -8 is not between -2 and 18 so it can’t be

so x=4

remember, y=-1/2x+3

sub that to get y=-1/2(4)+3=-2+3=1

the point is (4,1)

apologies for mishap

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The frequency of light that has a wavelength of 249 nm is about **1.20 × 10¹⁵ sec⁻¹**

### Further explanation

The term of package of electromagnetic wave radiation energy was first introduced by **Max Planck**. He termed it with **photons** with the magnitude is :

*E = Energi of A Photon ( Joule )*

*h = Planck’s Constant ( 6.63 × 10⁻³⁴ Js )*

*f = Frequency of Eletromagnetic Wave ( Hz )*

**The photoelectric effect** is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.

*E = Energi of A Photon ( Joule )*

*m = Mass of an Electron ( kg )*

*v = Electron Release Speed ( m/s )*

*Ф = Work Function of Metal ( Joule )*

*q = Charge of an Electron ( Coulomb )*

*V = Stopping Potential ( Volt )*

Let us now tackle the problem !

__Given:__

λ = 249 nm = 2,49 × 10⁻⁷ m

__Unknown:__

f = ?

__Solution:__

### Learn more

### Answer details

**Grade: **College

**Subject: **Physics

**Chapter:** Quantum Physics

**Keywords:** Quantum , Physics , Photoelectric , Effect , Threshold , Wavelength , Stopping , Potential , Copper , Surface , Ultraviolet , Light

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From top to bottom in a group of elements, ionization energies tend to… A) remain constant. B) increase. C) decrease. D) alternate.

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**Answer:**

The initial height is 5 feet.

The object will hit the ground after approximately 4.57 seconds.

**Step-by-step explanation:**

An object is launched into the air. The projectile motion of the object can be modeled using the function h(t) = -16t^2 + 72t + 5, where t is the time in seconds since the launch and h(t) represents the height in feet of the object after t seconds

General equation is

V_0 is the initial velocity

h_0 is the initial height

**From the given equation , the initial height is 5 feet**

Initial velocity is +72 feet / sec

When the onbject hits the ground, the height becomes 0

So we plug in 0 for h(t) and solve for t

USe quadratic formula to solve for t

a=-16, b= 72, c= 5

t= -0.06 and t= 4.568

**The object will hit the ground after approximately 4.57 seconds.**

To find out the height after 3 seconds, plug in 3 for t

At t=0 , h(0) = 5

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Help me with the bottom part please!

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**Answer:**

A cross section parallel to the base of a right rectangular prism is congruent to the base.

**Step-by-step explanation:**

1. A cross section parallel to the base of a right rectangular prism is a square. This is false as the cross section will be a rectangle if the base is rectangular and square if the base is a square.

2. A cross section perpendicular to the base of a right rectangular prism is congruent to the base. No, only parallel cross sections are similar to base.

3**. A cross section parallel to the base of a right rectangular prism is congruent to the base. This is true** as a cross section parallel to any shape will give the shape of the base. With each slice, the size gets smaller but the shape is the same as base.

4. A cross section perpendicular to the base of a right rectangular prism has the same dimensions as the base. This is false as a completely different shape is generated with different dimensions.

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births

deaths

immigration

emigration

A population gains individuals by birth and immigration and loses individuals by death and emigration.

Biotic Potential

Populations vary in their capacity to grow. The maximum rate at which a

population can increase when resources are unlimited and environmental

conditions are ideal is termed the population’s biotic potential. Each

species will have a different biotic potential due to variations in

the species’ reproductive span (how long an individual is capable of reproducing)

the frequency of reproduction (how often an individual can reproduce)

“litter size” (how many offspring are born each time)

survival rate (how many offspring survive to reproductive age)

There are always limits to population growth in nature. Populations

cannot grow exponentially indefinitely. Exploding populations always

reach a size limit imposed by the shortage of one or more factors such

as water, space, and nutrients or by adverse conditions such as disease,

drought and temperature extremes. The factors which act jointly to

limit a population’s growth are termed the environmental resistance. The interplay of biotic potential and density-dependent environmental resistance keeps a population in balance.

Carrying Capacity

For a given region, carrying capacity is the maximum number of

individuals of a given species that an area’s resources can sustain

indefinitely without significantly depleting or degrading those

resources. Determining the carrying capacities for most organisms is

fairly straightforward. For humans carrying capacity is much more

complicated. The definition is expanded to include not degrading our

cultural and social environments and not harming the physical

environment in ways that would adversely affect future generations.

For populations which grow exponentially, growth starts out slowly,

enters a rapid growth phase and then levels off when the carrying

capacity for that species has been reached. The size of the population

then fluctuates slightly above or below the carrying capacity.

Reproductive lag time may cause the population to overshoot the carrying

capacity temporarily. Reproductive lag time is the time required for

the birth rate to decline and the death rate to increase in response to

resource limits. In this scenario, the population will suffer a crash

or dieback to a lower level near the carrying capacity unless a large

number of individuals can emigrate to an area with more favorable

conditions. An area’s carrying capacity is not static. The carrying

capacity may be lowered by resource destruction and degradation during

an overshoot period or extended through technological and social

changes.

An example of dieback occurred in Ireland after a fungus infection

destroyed the potato crop in 1845. During this potato famine

approximately 1 million people died and 3 million people emigrated to

other countries. Increased food production due to improved agricultural

practices, control of many diseases by modern medicine and the use of

energy to make historically uninhabitable areas of Earth inhabitable are

examples of things which can extend carrying capacity. The question is

how long will we be able to keep increasing our population on a planet

with finite size and resources?

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In Wordsworth’s poem nature is powerful and inviting, exhibiting forces of healing in the form of bright colors and gentle vibes. It is recounted from a comfortable, safe perspective; when the speaker is resting on his safe, warm couch, the memories of his solo walk along the bay

…flash upon that inward eye

Which is the bliss of solitude;

And then my heart with pleasure fills,

And dances with the daffodils.

These recollections serve as a comfort and pleasure to him, even when he is comfortable in a pleasant environment. Such was the power of the scene.

De la Mare’s poem also presents nature as a powerful force, but an impersonal, destructive one. The poem is told from the perspective of sea birds in a storm, and the vocabulary is a violent as Wordsworth’s is serene: “And the wind rose, and the sea rose,/To the angry billows’ roar,” and in the second verse,

And the yeasty surf curdled over the sands,

The gaunt grey rocks between;

And the tempest raved, and the lightning’s fire

Struck blue on the spindrift hoar –

Here the birds have lost control, and the storm is forcing them onto the shore, waves tossing and wind howling, a wholly different scene than Wordsworth’s happy spring day. Even in the end, when the storm breaks and the sun comes out, we see the lingering effects of the chaos – “the bright green headlands shone/As they’d never shone before,” and yet within this setting we have vast hoards of sea birds breaking this lovely post-storm calm with their “screeching, scolding, [and] scrabbling.” But in the final two lines of the poem, we see also “A snowy, silent, sun-washed drift/Of sea-birds on the shore.” And herein lies the true destruction: while a whole host of birds are tumbling through the sky, another host of birds has been killed by the violence of the storm.

Both poems depict the unpredictability of nature, and yet because Wordsworth’s poem is from the point of view of a man, on a bright spring day, his poem is more domestic and simple than that of de la Mare. The latter presents the point of view of nature itself, only to switch to a third person, withdrawn perspective at the end of the poem; humans have no role in the events that unfold. Any humans that exist in the area would have been safely indoors during the storm, away from any danger. We therefore get the rawness of nature where we would normally escape it for our fires and our beds; here is the flip-side of natural beauty – natural destruction. This poem is no walk in the garden, but a story of the wildness of natural processes.

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