Questions on Word Problems: Geometry answered by real tutors!

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L=2W+5
2L+2W=82
2(2W+5)+2W=82
4W+10+2W=82
6W=82-10
6W=72
W=72/6
W=12 IS THE WIDTH.
L=2*12+5
L=24=5
L=29
PROOF:
2*29+2*12=82
58+24=82
82=82

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The longer leg of a right triangle is 4 cm more than four times the length of the shorter leg. The hypotenuse is 1 cm longer than the longer leg. How long are the sides of the triangle?
.
Let x = length of shorter leg
then
4x+4 = length of longer leg
4x+5 = length of hypotenuse
.
x^2 + (4x+4)^2 = (4x+5)^2
x^2 + (4x+4)(4x+4) = (4x+5)(4x+5)
x^2 + 16x^2+32x+16 = 16x^2+40x+25
17x^2+32x+16 = 16x^2+40x+25
x^2+32x+16 = 40x+25
x^2-8x+16 = 25
x^2-8x-9 = 0
(x-9)(x+1) = 0
.
x = {9, -1}
We can toss out the negative solution leaving:
x = 9 cm (shorter leg)
.
Longer leg:
x+4 = 9+4 = 13 cm
.

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L=W+25
2(L-2*2)(W-2*2)=372
2(W+25-4)(W-4)=372
2(W+21)(W-4)=372
W^2+21W-4W-84=372/2
W^2+17W-84=186
W^2+17W-84-186=0
W^2+17W-270=0
(W+27)(W-10)=0
W-10=0
W=10 ANS. FOR THE WIDTH.
L=10+25
L=35 ANS. FOR THE LENGTH.
PROOF:2(35-4)(10-4)=372
2*31*6=372
372=372

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We know , working eqn
But, , subst in our working eqn





Also,
In doubt? Go back working eqn,



Thank you,
Jojo

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26^5 = 11,881,376
====================
Cheers,
Stan H.

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Originally the dimensions of a rectangle were 20 cm by 23 cm. When both dimensions were decreased by the same amount, the area of the rectangle decreased by 120 cm^2. Find the dimensions of the new rectangle.
---------------------------
Original DATA:
area = 20*23 = 460 cm^2
--------------------------
New DATA:
area = (20-x)(23-x) cm^2
------------------------------
EQUATION:
original area - new area = 120 cm^2
460 -[460-43x+x^2] = 120
460 - 460 +43x - x^2 = 120
x^2 - 43x + 120 = 0
x = [43 +- sqrt(43^2 -4*120)}/2
x = [43 +- sqrt(1369)]/2
x = [43 +- 37]/2
Positive solution:
x = 3 cm or x = 40 cm
Only x = 3 cm is realistic.
=============================
Cheers,
Stan H.

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20*23=460cm^2 is the original area.
(20-x)(23-x)=460-120
460-43x+x^2=340
x^2-43x+460-340=0
x^2-43x+120=0
(x-40)(x-3)=0
x-3=0
x=3
Thus the new dimentions are:
20-3=17
23-3=20
Proof:
17*20=460-120
340=340

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what is the answer to the problem 2x+2
---------------
2x+2 is not a problem. It's just an expression, a binomial.

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If a regular Hexogram (6 sides) has a overall diameter of 10', what would each of the 6 sides measure?
-----------------
one side = 10'/6 = 5/3 = 1 2/3 ft
===================================
Cheers,
Stan H.

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let x be the number of rows and y be the number of trees in each row.
:
x*y=84...eq 1
x=y+5....eq 2
:
take x's value from eq 2 and plug it into eq 1
:
(y+5)y=84
:

:
solve by quadratic formula y=7 and y =-12....throw out the negative value
:
using y's value in eq 2 we get:rows
:

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Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=361 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 7, -12. Here's your graph:

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"cross" multiplying __ L^2-LW=W^2 __ dividing by W^2 __ (L/W)^2-(L/W)=1

(L/W)^2-(L/W)-1=0 __ using quadratic formula __ (L/W)=[1±sqrt(5)]/2

negative value is not realistic so L/W=[1+sqrt(5)]/2 __ this is the "Golden Ratio"

W=2L/(1+sqrt(5)) __ W=(2*36)/(1+sqrt(5)) __ W=22.25 (approx)


#84 __ working in miles and hours, so 20min is 1/3 hr __ t=d/r

let x="rate of boat" __ t=5/(x+4) __ t+(1/3)=5/(x-4)

substituting __ [5/(x+4)]+(1/3)=5/(x-4)

multiplying by 3(x+4)(x-4) __ 15x-60+x^2-16=15x+60

subtracting 15x+60 __ x^2-136=0 __ x=sqrt(136) __ x=11.66 (approx)

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Can you tell me what the perimeter of a rectangular backyard is that is
6x + 6 yards. If the width is x yards, can you please find a bionomial that
represents the length. Thanks!

The perimeter is 2 times the width plus 2 times the length. If the width is x and the perimeter is 6x+6, than the expression that equals the length would be:



The expression above is derived from the following:





Now simplify the expression :

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Find the exact area of a triangle with a base of square root 30 meters and a height of square root 6 meters.
-----------------------
The area is bh/2, 1/2 of base times height.
For this one, it's:

=
=
=
=

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Hello there, this is the answer for this problem, perhaps more algebraic than geometrical:
We have the largest angle designated as A, the second largest as B and the smallest as C.
Every step must be notated in order to get the equations.
We know that the largest angle is twice the second largest which means: A=2B
Also we know that one-third of the largest angle is 10 degrees larger than the difference of the other two, so, we have: (1/3)A=B-C+10
As our last equation, we know that the sum of all interior angles in a triangle is 180, that is: A+B+C=180
We have our three equations:
A=2B
(1/3)A=B-C+10
A+B+C=180
From there we solve them, results are:
A=102
B=51
C=27
Which, in this case, the one we need is the smallest angle, that is C, which is 27 degrees.

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a = largest angle
b = second largest angle
c = smallest angle
-----
a = 2*b
(1/3)*a = b - c + 10
-----
c formula becomes:
c = b - (1/3)*a + 10
which becomes:
c = b - (2/3)*b + 10
which becomes
c = (1/3)*b + 10
-----
a + b + c = 180 (sum of angles of a triangle = 180)
2*b + b + (1/3)*b + 10 = 180
2*b + b + (1/3)*b = 170
3*b + (1/3)*b = 170
multiply both sides by 3:
9*b + b = 510
10*b = 510
b = 51
-----
2*b = 102 = a
c = b - (1/3)*a + 10
c = 51 - (1/3)*102 + 10
c = 27
-----
a = 102
b = 51
c = 27
-----
a + b + c = 180
102 + 51 + 27 = 180
180 = 180
ok
-----
a = 2 * b
102 = 2 * 51
102 = 102
ok
-----
(1/3)*a = (b-c) + 10
1/3)*102 = (51 - 27) + 10
34 = 30
ok
-----

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We know that , EQN 1
But , exceeds the base by 4 inches
So, , cross multiply

, factor out being perfect square

Therefore, ;


Check,



Thank you,
Jojo

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x^2+(x+50)^2=250^2
x^2+x^2+100x+2,500=62,500
2x^2+100x+2,500-62,500=0
2x^2+100x-60,000=0
2(x^2+50x-30,000)=0
2(x-150)(x+200)=0
x-1150=0
x=150 for one side.
150+50=200 for the other side.
Proof:
150^2+200^2=250^2
22,500+40,000=62,500
62,500=62,500

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This word problem involves quadratic equations which I am having a very difficult understanding. The width of a rectangle is 4 ft less than the length. The area is 12ft^2. Find the length and width. I start out translating the equation like this:
A=LW
L=length
4-L=width <<--SHOULD BE L-4
.
Let x = length
then
x-4 = width
.
12 = x(x-4)
12 = x^2-4x
0 = x^2-4x-12
Factoring:
0 = (x-6)(x+2)
x = {-2, 6}
.
A negative solution doesn't make sense -- so, toss it out.
x = 6 feet (length)
.
Width:
x-4 = 6-4 = 2 feet (width)

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A) V(10)=10^3-13(10)^2+54(10)-72=168
B) you have to factor V(S)
C) is not right your answer, here the idea is to use B)

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29^2+x^2=(x+1)^2
841+x^2=x^2+2x+1
2x+1-841=0
2x=841-1
2x=840
x=840/2
x=420 for the other side.
420+1=421 is the length of the hypotenuse.
Proof:
29^2+420^2=421^2
841+176,400=177,241
177,241=177,241

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-eq where we will call a the shortest leg, b the other leg,and c the hypothenuse. Lets call b=b and c=b+1 where b and b+1 are two consecutive whole numbers. no substitute our known numbers and variables into the equation.
solving we have
other leg hypothenuse

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let 'l' be the length and 'w' be the width of the flag.
then perimeter = 2l +2w ( rectangle)
given, perimeter = 28 ft
i.e. 2l +2w = 28 ft
2 (l +w) = 28
l + w = 28 /2 = 14 ..........(1)
given length is 4 ft less than its width
i.e . l = w - 4 ...........(2)
from (1) & (2)
w - 4 + w = 14
2w - 4 = 14
2w = 14+4
2w = 18
w = 18 /2 = 9ft
l = w - 4 = 9 - 4 = 5 ft

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P=2w+2l eq 1
l=2w-4 eq 2
:
substitute P=28 and l's value in eq 2 into eq 1--->28=2w+2(2w-4)
:
28=2w+4w-8---- so

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tan 25=24000/y where y is the distance between a point on the ground below
:
: the plane and the airport---->24000=tan 25 (y)----->y=51,502ft
:
sin 25=24000/x where x is the distance between the plane and the airport
:
: 24000=sin 25(x)-----y=56,738ft

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A flat panel monitor are such that its length is 3 in. more than its width. If length were doubled and if the width were decreased by 1 in., the area would be increased by 150 in.2(squared). What are the length and width of flat panel monitor?
.
Let w = width of monitor
then
w+3 = length of monitor
.
double length = 2(w+3)
width decreased by 1 = w-1
.
2(w+3)(w-1) =150
2(w^2-w+3w-3) =150
2(w^2+2w-3) =150
w^2+2w-3 =75
w^2+2w-78 = 0
.
Using the quadratic equation to solve, we find the roots as:
w = {7.888, -9.888}
.
We can toss out the negative solution since it doesn't make sense.
w = 7.888 inches (width)
.
length:
w+3 = 7.888+3 = 10.888 inches (length)
.
Details of quadratic:

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Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=316 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 7.88819441731559, -9.88819441731559. Here's your graph:

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To earn money, his profits have to be positive.

Let's find the point where the profits equal zero.
From that point on, his profit will be positive.



.
.
.
Check the equation.
John makes a $79 profit selling 0 hot dogs, that can't be right.
Also, when he sells more than a certain number (63), his profits become negative, that doesn't make sense either.
Check and re-post the problem.

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1st one: A lot is in the shape of a right triangle. The shorter leg measures 180 m. The hypotenuse is 60 m longer than the length of the longer leg. How long is the longer leg??
.
Let x = length of longer leg
then
x+60 = hypotenuse
.
Applying pythagorean's theorem:
180^2 + x^2 = (x+60)^2
180^2 + x^2 = (x+60)^2
32400 + x^2 = x^2 + 120x + 3600
32400 = 120x + 3600
28800 = 120x
240 meters = x (longer leg)
.
2nd one: A rectangular piece of cardboard measuring 28 inches by 29 inches is to be made into a box with an open top by cutting equal size squares from each corner and folding up the sides. Let x represent the length of a side of each such square. For what value of x will the volume be a maximum? If necessary, round to 2 decimal places.
.
Volume = x(28-2x)(29-2x)
Volume = x(812-56x-58x+4x^2)
Volume = x(812-114x+4x^2)
Volume = 812x-114x^2+4x^3
Volume = 4x^3-114x^2+812x
.
Taking the derivative:
4x^3-114x^2+812x
12x^2-228x+812 = 0
6x^2-114x+406 = 0
Solving the above with the quadratic equation we get:
x = {14.25, 4.75}
.
Solution: 4.75 inches
.
Details of quadratic equation:

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Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=3252 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 14.2521924764611, 4.74780752353892. Here's your graph:

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"A square and a rectangle have the same perimeter" __ 4S=2L+2W

"The lenghth of the rectangle is 4cm less than twice the side of the square" __ L=2S-4

"the width of the rectangle is 6cm less than the side of the square" __ W=S-6

substituting __ 4S=2(2S-4)+2(S-6) __ 4S=4S-8+2S-12 __ 4S=6S-20 __ 20=2S __ 10=S

so the perimeter is 4*10 or 40

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Let x be the altitude, then the base is x + 9.
Its area is
As its area is 56, we have

Solving for x, we have






So

Or
(reject this negative root)
Thus the length of the altitude is 7, and the length of the base is x+9 = 7+9=16.

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so we will call the shorter side s and longer side is 2s+2 . Hypothenuse is 13
using the pathagorean theorem multiplied out we get 5s2+8x-165=0
factoring we get (5s+33)(s-5)=0 so s=5 and -33/5 throw out the negative value and s=5 meaning short side is 5 and longer side is 2(5)+2=12

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A collection of triangles and pentagons contains 40 shapes having a total of 138 sides. How many pentagons are there?
----------
# of shapes equation: p + t = 40
# of sides equation: 5p +3t = 138
--------------------------
Multiply thru 1st by 3 to get:
3p + 3t = 120
5p + 3t = 138
----------------
Subtract 1st from 2nd to get:
2p = 18
p = 9 (# of pentagons)
Since p + t = 40, t = 31 (# of triangles)
=====================
Cheers,
Stan H.

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a) Remember:
But, , length 4 ft more than its width
Then,

--->

So,
Check,


, good
b) Remember:

Take note:
Then,
=
= ----> cost to carpet the whole room
Thank you,
Jojo

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so we are given P-perimeter as 52 feet we know that P=2(L+W) the length or L is 4 more fee than the width L=W+4 so substitution our given and the L value in the 2nd equation into our perimeter formula we get 52=2((W+4)+W) lets distribute the right side 52=4W+8 so subtract 8 both sides and 4W=44 divide each side by 4 and W=11 L=W(11)+4 so L=15
a. width is 11 feet
length is 15 feet
b A=15 times 11=165 sq ft but we must divide this by 9 to get square yards
so 165/9=18.33 sq yards
so 18.33 sq yd multiplied by cost per yard of $18.95 equals $347.35 as being the cost of the carpet for this given room

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---> 24ft &
---> , ANSWER
Thank you,
Jojo

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L*W = 16
K*L*K*W = 64
since a*b = b*a, and (a)*(b*c) = (a*b)*(c), this equation can become
(K*K)*(L*W) = 64
since L*W = 16, this equation can become
(K*K)*(16) = 64
divide both sides by 16
K*K = 64/16 = 4
K*K = 4
K = 2
-----
your scaling factor is 2.
-----
you multiply each dimension by 2 which means the whole thing is multiplied by 4.
-----

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The width of a rectangle is 3 feet less than its length. if the perimeter is 22 feet, what are both its length and width?
width = x - 3
length = x
perimeter = 22
P = 2L + 2W
22 = 2x + 2(x - 3)
22 = 2x + 2x - 6
22 = 4x - 6
22 + 6 = 4x
28 = 4x
28/4 = x
7 = x
Since x = length, then length is 7 feet.
Our width is x - 3.
Replacing x with 7 and subtracting 3, will give us 4 feet.
Final answer: width = 4 feet and length = 7 feet.

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P=2(L+W) W=L-3 substituting W and P(given as 22ft) into the first equation we get 22=2(L+(L-3) so 22=2(2L-3) distributing the right side of the equation we arrive at 22=4L-6. Add 6 and divide by 4 on each side of the equation and we get L=7 since L=7 then W=L(7)-3 so W=4
L=7
W=4

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86*42=3,612 ft^2 area of the plot
(86+6)(42+6)=92*48=4,416 ft^2 area of the plot + the walk.
4,416-3,612=804 ft^2 is the area of the walk.

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5x+5(x+4)=520
5x+5x+20=520
10x=520-20
10x=500
x=500/10
x=50 mph for the slower car.
50+4=54 mph for the faster car.
Proof:
5*50+5*54=520
250+270=520
5290=520

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.20x+.50(8-x)=.30*8
.20x+4-.50x=2.4
-.30x=2.4-4
-.30x=-1.6
x=-1.6/-.30
x=5.333 pints of 20% are used.
8-5.333=2.667 pints of 50% are used.
Proof:
.20*5.333+.50*2.667=2.4
1.0666+1.333=2.4
2.4=2.4

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.08x=.10(5400-x)
.08x=540-.10x
.08x+.10x=540
.18x=540
x=540/.18
x=$3,000 invested @ 8%.
5,400-3,000=$2,400 invested @ 10%.
Proof:
.08*3,000=.10*2,400
240=240

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The problem is - a beach ball holds 800 cubic inches of air. Another beach ball has a radius that is half of the larger ball. How much air doess the smaller ball hold? I know the formula for finding volume for a sphere - but I don't know how to use it without any additional information. Thanks for your help in this.
-------------------------

You could solve for the radius, then find the volume of the smaller ball, but that's not necessary.
The volume is a function of the cube of the radius. If the radius is doubled, the volume is 2^3 time, or 8 times.
The 2nd ball's radius is 1/2, so its volume is (1/2)^3 times, or 1/8.
So it's 100 cubic inches or air.

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Thank you,
Jojo

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Call the smaller leg
Call the hypotenuse
The longer leg =





You can solve by completing the square



Take the square root of both sides




The sides are 10,24, and 26
check answer




OK

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Let x=amount that needs to be drained off and replaced with pure antifreeze
Now we know that the amount of pure antifreeze left after x amount is drained off (0.4(50-x)) plus the amount of pure antifreeze added(x) has to equal the amount of pure antifreeze in the final mixture (0.5(50)). So, our equation to solve is:
0.4(50-x) + x=0.50*50 get rid of parenthesis(distributive)
20-.4x+x=25 subtract 20 from each side
and combine like terms
.6x=5 divide each side by 0.6
x=8.33 qts-----------------------------------ans
CK
.4(50-8.33)+8.33=25
(50/3)+25/3=25
75/3=25
25=25
Hope this helps