course mth 152
I was wondering if you need me to retake the tests on the previous chapters. On my grade page it says to email you and do some more work on query assignments. What do I need to work on? Thanks for your timeConnie
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution.
If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
023. ``q Query 23
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Self-critique (if necessary):
Self-critique Rating:
Question: **** `q Query 9.4.6 ABC, DEF transversed by EOB at rt angles; OB = EO; show triangles ABO and DOF congruent.
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Your solution:
If OB= EO then EO= OB angle BOA= angle DOE because they are vertical
so triangle ABO= triangle DOE because they are side angle congruent
Confidence Assessment:
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Given Solution:
`a SAS: Angle AOB and Angle FOE are equal because they are vertical angles, so we have 2 sides and the included angle of triangle AOB equal, respectively, to 2 sides and the included angle of triangle FOE. Thus, the Side-Angle-Side property holds that triangle AOB is congruent to triangle FOE.
**** Explain the argument you used to show that the triangles were congruent.
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Self-critique (if necessary):
Self-critique Rating:
Question: **** `q Query 9.4.18 ACB and QPR similar triangles, C and P rt angles, A=42 deg **** List the measures of the three angles of each triangle and explain how you obtained each.
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Your solution:
Angle C =90 because it is right angle Angle P = 90 because it is right angle
Angle A = 42 Angle B = 180- (90+42) = 180-132= 48 Angle B=48
since the triangles are similar then Angle Q= Angle A so 42 is Angle Q
Angle B= Angle R so R is 48
Angle c= 90 Angle a= 42 Angle B= 48
Angle P = 90 Angle Q= 42 Angle R= 48
Confidence Assessment:
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Given Solution:
`a It is given that Angle A = 42 deg. and Angle C = 90 deg. Since all three angles must add up to equal 180 then Angle B = 48 deg.
In the second triangle, Angle P must equal 90 deg. since it is a right angle.
To find Angle R,
90(48) = 90R sp
4320 = 90R and
48 = R Angle R = 48 deg.
To find Angle Q,
90/90 = Q/42
Q = 42
Angle Q = 42 deg.
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Self-critique (if necessary):
Self-critique Rating:
Question: **** `q Query 9.4.24 similar triangles, corresp sides a, b, 75; 10, 20, 25 **** What are the lengths of sides a and b and how did you obtain each?
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Your solution:
If C= 75 and A is unknown and B is unknown but you know D E F is 10 20 and 25 since they are similar then C is similar to F so 75 is 3 times 25 so the sides are triple what you know
C=75 B= 60 because 20*3 A= 30 because 10*3 D= 10 E= 20 F = 25
Confidence Assessment:
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Given Solution:
`a To find a,
75 (10) = 25a
750 = 25a
a= 30
To find b,
75/25 = b/20
1500/25 = 25b/25 so
b = 60.
a = 30, b = 60 and c = 75.
These values are triple the values of the similar triangle.
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Self-critique (if necessary):
Again I dont remember the formula you used but I understand the answer
Self-critique Rating:
Question: **** `q Query 9.4.42 rt triangle a = 7, c = 25, find b **** What is the length of side b and how did you obtain it?
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Your solution:
if you have a right triangle then you use the Pythagorean theorem a^2+ b^2 =c^2 c is your hypotenuse
7^2 + b^2= 25^2
49+ b^2 = 625
b^2= 576
b=24
Confidence Assessment:
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Given Solution:
`a By the Pythagorean Theorem a^2 + b^2 = c^2. So we have
49 + b^2 = 625 Subtract 49 from both sides to get
b^2 = 576. Take the square root of both sides to get
b = 24.
**** What does the Pythagorean Theorem say about the triangle as given and how did you use this Theorem to find the length of b?
Student Response: It says the sum of the squares of the lengths of the legs is equal to the square of the hypotenuse. I showed that this is true in the previous problem. I squared the legs and they equaled the hyppotenuse squared.
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Self-critique (if necessary):
I undestand the Pythageorean theorem
Self-critique Rating:
Question: **** `q Query 9.4.60 m, (m^2 +- 1) / 2 gives Pythagorean Triple **** What Pythagorean Triple is given by m = 5?
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Your solution:
if you have a triple it stil nedds to be a^2+ b^2 = c^2
m=5
( 5^2+1) / 2 = (25+1) /2 = 13
(5^2-1) /2= 24/2 =12
so you have 5, 12, 13 you should have 5^2+ 12^2= 13^2
25+ 144= 169
169 = 169
Confidence Assessment:
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Given Solution:
`a ** If m = 5 then
(m^2 + 1) / 2 = (5^2 + 1 ) / 2 = 26 / 2 = 13
(m^2 - 1) / 2 = (5^2 - 1 ) / 2 = 24 / 2 = 12
So the Pythagorean triple is 5, 12, 13.
We can verify this:
5^2 + 12^2 should equal 13^2.
5^2 + 12^2 = 25 + 144 = 169.
13^2 = 169.
The two expressions are equal so this is indeed a Pythagorean triple. **
**** How did you verify that your result is indeed a Pythagorean Triple?
Student Answer: The numbers checked out when substituted into the Pythagorean Theorem.
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Self-critique (if necessary):
Self-critique Rating:
Question: **** `q Query 9.4.75 10 ft bamboo broken, upper end touches ground 3 ft from stem. **** How high is the break, and how
did you obtain your result?
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Your solution:
if the tree breaks and it was 10 feet before the break then the long side or hypotenuse is 10-x and x is
the tree up to the break and the base is 3 because it says there is 3 feet of tree on the ground
x^2 + 3^2 = (10-x)^2 10 - 4.55 = 5.45 is the broken part
x^2+ 9= (10-x) * (10- x)
x^2 +9 = 100 -10x -10x +x^2
x^2 +9 = 100 -20x + x^2
9= 100 - 20 x
-91 = -20x
x= 4.55
Confidence Assessment:
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Given Solution:
`a ** If the break is at height x then the hypotenuse, consisting of the broken part, is at height 10 - x.
The triangle formed by the vertical side, the break and the ground therefore has legs x and 3 and hypotenuse 10-x.
So we have
x^2 + 3^2 = (10-x)^2. Squaring the 3 and the right-hand side:
x^2 + 9 = 100 - 20 x + x^2. Subtracting x^2 from both sides
9 = 100 - 20 x so that
-20 x = -91 and
x = 4.55.
The break occurs at height 4.55 ft and the broken part has length 10 - 4.55 = 5.45, or 5.45 feet. **
**** How did the Pythagorean Theorem allow you to solve this problem?
I substituted the numbers into the Pythagorean Theorem.
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Self-critique (if necessary):
Self-critique Rating:
Question: **** `q Query 9.4.84 isosceles triangle perimeter 128 alt 48 **** What is the area of the triangle and how did you find
it?
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Your solution:
In an isosceles triangle 2 sides are equal so if you have triangle abc then side a= side b and c is the
base we know that the perimeter is 128 so that would mean that c= 128- side b and side a or c=128-2x
so if you take the altitude and draw a line down to the base it would make a side to a new triangle of 48
with a right angle and then the length of c would be 1/2(128-2x) or 64-x and b would be the same or x
so take the new side of 48 and solve 48^2+ (64-x)^2 = x^2
2304 + (64-x) *(64-x) =x^2
2304+ 4096 - 64x -64x +x^2 =x^2
2304+4096- 128x+x ^2 =x^2
6400 -128x +x^2 = x^2
-128x= -6400
x=50
so if a=b then side a =50 side b = 50 and side c = 128-(2*50) or 128 - 100 or 28
area = 1/2 b*h 1/2 (28*48) = 1/2(1344) = 672
Confidence Assessment:
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Given Solution:
`a ** This problem is algebraically demanding. Your text might have a slicker way to do this, but the following works:
If the equal sides are x then the base is 128 - 2 x.
The altitude forms a right triangle with half the base and one of the equal sides. The sides of this right triangle are therefore 48, 1/2 (128 - 2x) = 64 - x, and x.
The right angle is formed between base and altitude so x is the hypotenuse.
We therefore have
48^2 + (64 - x)^2 = x^2 so that
48^2 + (64 - x) ( 64 - x) = x^2 or
48^2 + 64 ( 64-x) - x(64 - x) = x^2 or
48^2 + 64^2 - 64 x - 64 x + x^2 = x^2 or
48^2 + 64^2 - 128 x + x^2 = x^2. Subtracting x^2 from both sides we get
48^2 + 64^2 - 128 x = 0. Adding 128 x to both sides we get
48^2 + 64^2 = 128 x. Multiplying both sides by 1/128 get have
(48^2 + 64^2) / 128 = x. Evaluating this expression we end up with x = 50.
The base of the triangle is therefore 128 - 2x = 128 - 2 * 50 = 128 - 100 = 28.
So its area is 1/2 b h = 1/2 * 28 * 48 = 672. **
DRV
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Self-critique (if necessary):
Self-critique Rating:
Question: **** `q Query Add comments on any surprises or insights you experienced as a result of this assignment.
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If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution.
If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
024. ``q Query 24
Question: **** `q Query 9.5.12 vol of sphere diam 14.8 **** What is the volume of the sphere and how did you obtain it?
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Your solution:
the formula for a sphere volume is V=4/3 pi r^3
so if the diameter is 14.8 then the radius is 7.4
V= 4/3*3.14* (7.4)^3
4/3* 3.14* 405.224
4/3*1272.40
1696.54
Confidence Assessment:
*********************************************
Given Solution:
`a I use the formula for finding the volume of a sphere which is 4/3(3.14)(r^3).
Since the diameter is 14.8, the radius is half that which is 7.4.
V = 4/3 * 3.14 * 7.4^3
V = 4/3 * 3.14 * 405.224
V = 1696.54
The volume is 1696.54
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Self-critique (if necessary):
Self-critique Rating:
Question: **** `q Query 9.5.18 pyramid 12 x 4 altitude 10 **** What is the volume of the pyramid and how did you find it?
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Your solution:
volume of a pyramid is V= 1/3 Bh
the base is a square so 12*4 =48
V=1/3 48*10
V= 1/3* 480
V= 160
Confidence Assessment:
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Given Solution:
`a I used the formula : V = 1/3Bh
The base = 12 * 4 = 48
V = 1/3 * 48 * 10
V = 1/3 * 480
V = 160
The volume is 160ft.^3
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Self-critique (if necessary):
Self-critique Rating:
Question: **** `q Query 9.5.24 bottle 3 cm alt 4.3 cm **** What is the volume of the bottle and how did you find it?
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Your solution:
I think the bottle would be a cylinder if so then V= pi r^2 h diameter is 3 so radius is 1.5
V= 3.14* (1.5)^2* 4.3
V= 3.14*2.25* 4.3
V= 7.065*4.3
V= 30.38 cm^3
Confidence Assessment:
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Given Solution:
`a ** The figure is a right circular cylinder with V = 3.14 * r^2 * h
Since the diameter is 3, then the radius is 1.5
V = 3.14 * 1.5^2 * 4.3
V = 3.14 * 2.25 * 4.3
V = 30.38 cm^3 **
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Self-critique (if necessary):
Self-critique Rating:
Question: **** `q Query 9.5.36 sphere area 144 `pi m^2 **** What are the radius, diameter and volume of the sphere and how did you find them?
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Your solution:
the sphere area is S=4 pi r^2 if the radius is 6 then the diameter is 12
4pir^2 = 144pi m^2
4 pi r^2/4 pi= 144 pi m^2/4 pi
r^2= 36 m^2
r=6 m
V= 4/3pi( 6)^3
4/3*3.14* 216
V= 4/3* 678.24
V= 904.32 m^3
Confidence Assessment:
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Given Solution:
`a ** Sphere area is 4 pi r^2, so we have
4 pi r^2 = 144 pi m^2. Dividing by 4 pi we get
r^2 = 36 m^2. Taking the square root of both sides we get
r = 6 m.
From this we find that the diameter is 2 * 6 m = 12 m and the volume is 4/3 pi * (6 m)^3 = 288 pi m^3. **
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Self-critique (if necessary):
I do not understand the volume part. How did you get 48 pi m^3? I know that the formula is V=4/3 pi r^3
Im not quite sure how you did this one.
The calculation, which I corrected above, should have read
4/3 pi * (6 m)^3 = 288 pi m^3.
Self-critique Rating:
Question: **** `q Query 9.5.48 cone alt 15 rad x vol 245 `pi **** What is the value of x and how did you find your result?
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Your solution:
volume= 1/3 pi R^2h 245pi= 1/3 pi r^2*15
769.3 = 1.05 r^2*15
769.3= 15.75 r^2
48.84= r^2
6.99= r
Confidence Assessment:
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Given Solution:
`a ** We have V = 1/3 pi r^2 h. To solve for r we multiply both sides by 3 / (pi * h) to get
3 V / (pi * h) = r^2 then take the square root to get
r = sqrt(3 V / ( pi * h). Substituting we get
r = sqrt( 3 * 245 / (3.14 * 15) ) = 3.9 approx. **
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Self-critique (if necessary):
I think I understand. You need to simplify the formula first and then try to solve and finally plug in the information given
Self-critique Rating:
Question: **** `q Query 9.5.51 plane intersects sphere passing 7 in from center forming circle with area 576 `pi **** What is the volume of the sphere and how did you obtain it?
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Your solution:
area= pi*r^2 576*pi radius is square root of 576 = 24
24+7=30 for the radius of the sphere
V= 4/3 pir^3 V= 4/3 (3.14) * (30)^3
V= 4.19* 2700
V= 113040
Confidence Assessment:
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Given Solution:
`a ** The circle does have radius sqrt(576 in^2) = 24 in. However that is not the radius of the sphere since the plane containing the circle passes 7 in from the center of the sphere. So the center of the circle is not the center of the sphere.
The center of the circle is 7 in from the center of the sphere. Note also that a line from the center of the sphere to the center of the circle will be perpendicular to the plane of the circle.
Thus if you start at the center of the sphere and move the 7 in straight to the center of the circle, then move along the plane of intersection (in any direction) for 24 in (at which point you encounter the rim of the circle, whose radius you recall is 24 in), then back to the center of the sphere you will have traced out a right triangle with legs 7 in and 24 in. The hypotenuse of the triangle is the radius R of the sphere.
So we have
R^2 = 7^2 + 24^2 = 625 and
R = 25.
The radius of the sphere is 25 in.
Its volume will therefore be 4/3 pi r^3:
V = 4/3 pi r^3
= 4/3 pi * (25 cm)^3
= 4/3 pi * 16000 cm^3 (approx.)
= 67000 cm^3, approx. **
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Self-critique (if necessary):
I dont understand how you drew a triangle inside of the circle. How did you know to do that instead of adding the 7 to the radius of the plane?
The triangle wasn't inside the circle. It was, however, inside the sphere.
Its 7-inch leg runs from the center of the sphere to the center of the circle.
Its 24 in leg runs from the center of the circle to any point on the circle. That point is also on the sphere.
A radius of the sphere runs from the center of the sphere to any point on the sphere. The hypotenuse of the right triangle runs from the center of the circle to a point on the sphere, so the hypotenuse is a radius of the sphere.
The key part of the explanation is quoted in the paragraph below. You can probably get by OK without completely understanding this (nothing this challenging on the test), but if you still have trouble with the given solution and want to get to the bottom of it, then tell me, phrase by phrase, what you do and do not understand in the above explanation and in this paragraph:
'Thus if you start at the center of the sphere and move the 7 in straight to the center of the circle, then move along the plane of intersection (in any direction) for 24 in (at which point you encounter the rim of the circle, whose radius you recall is 24 in), then back to the center of the sphere you will have traced out a right triangle with legs 7 in and 24 in. The hypotenuse of the triangle is the radius R of the sphere.'
Self-critique Rating:
Question: **** `q Query Add comments on any surprises or insights you experienced as a result of this assignment.
&#Good work. See my notes and let me know if you have questions. &#