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Posted on Jan 02, 2017

Find the scaling ratio

Perimeter of first pentagon is 12*5=60 cm

Scale factor (2nd/first)=140/60 =7/3

**Ratio of areas varies as the square of the scale factor for the lengths**

Area of second / area of first =(7/3)^2 =49/9

Area of second pentagon =**248 *(49/9) cm^2**

Finish the calculation.

Perimeter of first pentagon is 12*5=60 cm

Scale factor (2nd/first)=140/60 =7/3

Area of second / area of first =(7/3)^2 =49/9

Area of second pentagon =

Finish the calculation.

Jul 01, 2014 | Texas Instruments TI-83 Plus Calculator

The two pentagons are similar. If k is the scale factor for the lengths, the ratio of the areas is k^2.

Pent 1 side =12 cm

Pent 2 side =140/5 =28 cm

Scale factor (larger over smaller) is k=28/12=7/3

Area of Pent_2= (49/9)*Area of Pent_1

Your answer should be C.

Pent 1 side =12 cm

Pent 2 side =140/5 =28 cm

Scale factor (larger over smaller) is k=28/12=7/3

Area of Pent_2= (49/9)*Area of Pent_1

Your answer should be C.

Jun 29, 2014 | Office Equipment & Supplies

Solve the resulting quadratic equation

2X^2+3X-90=0

Discriminant: (3)^2-4(2)(-90)=729=(27)^2

Two roots

X_1=(1/4)*(-3+27)=6

X-2=(1/4)*(-3-27) =-(15/2), negative

Since the width must be positive, reject the negative root and keep X_1=6

**Width =6**

Length=2(6)+3=15

Check 6*(15)=90. Checks OK

2X^2+3X-90=0

Discriminant: (3)^2-4(2)(-90)=729=(27)^2

Two roots

X_1=(1/4)*(-3+27)=6

X-2=(1/4)*(-3-27) =-(15/2), negative

Since the width must be positive, reject the negative root and keep X_1=6

Length=2(6)+3=15

Check 6*(15)=90. Checks OK

Apr 23, 2014 | Office Equipment & Supplies

Yes, there is shortcut because this is right triangle, so you can use Pythagorean theorem (see picture).

If this was helpful please rate 4 thumbs :)

- Length of hypotenuse is square root of sum of squares of lengths of other two sides of triangle, which is equal to square root of 30^2+10^2=31.6 cm.
- Sin(a)=longer cathetus/hypotenuse=0.949 so a=arcsin(0.949)=71.6 degrees
- Finally b=90-a=18.4 degrees.

If this was helpful please rate 4 thumbs :)

Sep 05, 2011 | Texas Instruments TI-30XA Calculator

Area = length X width, and you know the area and the width. So,

4122 sq cm = length X 3 cm

length = 4122 sq cm / 3 cm

length = 1374 cm

(That is one long, skinny rectangle, isn't it?)

4122 sq cm = length X 3 cm

length = 4122 sq cm / 3 cm

length = 1374 cm

(That is one long, skinny rectangle, isn't it?)

Mar 02, 2011 | Office Equipment & Supplies

If we assume the width of the frame is X. Then the one dimension of the framed picture will be 2X + 20 and the other dimension of the picture will be 2X + 34. We know the total area of the framed picture should be 640 in^2. Therefore:

(2X + 10) * (2X + 34) = 640

If you multiply this out you get:

4X^2 + 88X + 340 = 640

Subtract 640 from both sides:

4X^2 + 88X - 300 = 0

Divide both sides by 4:

X^2 + 22X - 75 = 0

Use the quadratic equation to solve for X, with A=1, B=22, C=75

X = [-22 +/- sqr(22^2-4*(-75)] /2

This reduces to:

(-22 +/- 48) / 2

The negative solution doesn't make sense in this case. Therefore: X = (-22 + 28) /2 = 6/3 = 3

Therefore the width of the frame should be 3 inches.

(2X + 10) * (2X + 34) = 640

If you multiply this out you get:

4X^2 + 88X + 340 = 640

Subtract 640 from both sides:

4X^2 + 88X - 300 = 0

Divide both sides by 4:

X^2 + 22X - 75 = 0

Use the quadratic equation to solve for X, with A=1, B=22, C=75

X = [-22 +/- sqr(22^2-4*(-75)] /2

This reduces to:

(-22 +/- 48) / 2

The negative solution doesn't make sense in this case. Therefore: X = (-22 + 28) /2 = 6/3 = 3

Therefore the width of the frame should be 3 inches.

Feb 16, 2011 | SoftMath Algebrator - Algebra Homework...

1. Well this one is pretty simple. Since you are not supposed to trim the length trim the width instead.

to make the perimeter 0.4 m shorter cut the cloth by 0.2 m width the picture will make it clear.

2. let the length of rectangle be x cm

and breadth = 20 cm

now original perimeter = 2(x+20) cm

new length = x-30 cm

breadth remains same = 20 cm

new perimeter = 2(x-30+20) cm = 2(x-10) cm

According to question

new perimeter = old perimeter /2

2(x-10) = 2(x+20) /2

2x -20 = x + 20

x = 40 cm

hence length of longer side 40 cm.

If you want to directly comm ankitsharma.220@gmail.com

to make the perimeter 0.4 m shorter cut the cloth by 0.2 m width the picture will make it clear.

2. let the length of rectangle be x cm

and breadth = 20 cm

now original perimeter = 2(x+20) cm

new length = x-30 cm

breadth remains same = 20 cm

new perimeter = 2(x-30+20) cm = 2(x-10) cm

According to question

new perimeter = old perimeter /2

2(x-10) = 2(x+20) /2

2x -20 = x + 20

x = 40 cm

hence length of longer side 40 cm.

If you want to directly comm ankitsharma.220@gmail.com

Feb 13, 2011 | Vivendi Excel@ Mathematics Study Skills...

The diameter of the circle is the diagonal of the rectangle use Pythagoras this rectangle is a made from two 3,4,5 triangles, so the diameter is 5cm, the radius will be 2.5cm don't forget the BODMAS ordering rule (Brackets of Division Multiplication Addition and Subtraction) Also rounding to 2 decimal places. This solution is based on all four corners of the rectangle touching the circumference of the circle.

Area of a circle is P (22/7) i x Radius (squared ^2) ====> 22/7 x 2.5cm ^ 2 ====> 22/7 x 6.25cm

TOTAL AREA of CIRCLE = 19.64 cm^2

Area of the rectangle is length x width ====> 4 cm x 3 cm

TOTAL AREA of RECTANGLE = 12 cm ^2

The remaining portion is the circle area minus the rectangle 19.64-12 = 7.64 cm^2

SIMPLES brought to you by SpideRaY @ http://www.windowstipsclub.com

Area of a circle is P (22/7) i x Radius (squared ^2) ====> 22/7 x 2.5cm ^ 2 ====> 22/7 x 6.25cm

TOTAL AREA of CIRCLE = 19.64 cm^2

Area of the rectangle is length x width ====> 4 cm x 3 cm

TOTAL AREA of RECTANGLE = 12 cm ^2

The remaining portion is the circle area minus the rectangle 19.64-12 = 7.64 cm^2

SIMPLES brought to you by SpideRaY @ http://www.windowstipsclub.com

Jan 31, 2011 | Computers & Internet

The surface area of the outer surface is the circumference of the outside of the pipe times its length. So, if OR is the radius of the outside of the pipe and SO is the outer surface area,

SO = 2 * pi * OR * 14

Similarly, the surface area of the outer surface is SI = 2 * pi * IR *14 (if SI is the inner surface area and IR is the radius of the inside of the pipe.

The question states that the difference between the outside surface area and the inside surface area is 88 sq. cm:

SO - SI = 88 ; substituting:

(2 * pi * OR * 14) - (2 * pi * IR * 14) = 88 ; factoring

(2 * pi * 14) (OR - IR) = 88 ;dividing both sides by 2*pi*14

(OR - IR) = 88/(2 * pi) = 88/(2 * 3.14159*14) = 1.00

So, the outside radius is 1 cm more than the inside radius.

It's not clear if the volume stated is the volume of the metal in the pipe or the volume of air inside the pipe, so I will solve it both ways:

If volume of 176cc is of the air inside, the formula for this volume is 14 * pi * IR *IR

176 = 14 * 3.14159 * IR * IR ; dividing both sides by 14 * 3.14159

176/(14 * 3.14159) = IR * IR ; doing the arithmetic

4 = IR * IR ; taking the square root of both sides

sqrt(4) = IR

**IR = 2 cm**

substituting back into the first equation, the OR is 1cm more than the IR, so

**OR = 3 cm**

If volume of 176cc is of the iron in the pipe, the formula for that volume is the difference between the volume of the outside of the pipe and the volume of the inside of the pipe, or

(14 * pi * OR * OR) - (14 * pi * IR * IR) = 176 ; factoring

(14 * pi) ((OR * OR) - (IR * IR)) = 176 ; dividing both sides by 14 * pi

((OR * OR) - (IR * IR)) = 176/(14 * 3.14159) = 4

but, since OR is 1 cm more than IR (from above), we can substitute OR = IR + 1

and OR * OR = (IR + 1) * (IR + 1) = (IR*IR) + 2*IR +1

So, ((OR * OR) - (IR * IR)) = 4 becomes

(IR*IR) + 2*IR +1 - (IR*IR) = 4 ; simplifying (IRsquared - IRsquared = 0)

2*IR + 1 = 4 ; subtract 1 from both sides

2*IR = 3 ; divide both sides by 2

IR = 3/2 ;

**IR = 1.5cm**

**OR = 2.5cm**

SO = 2 * pi * OR * 14

Similarly, the surface area of the outer surface is SI = 2 * pi * IR *14 (if SI is the inner surface area and IR is the radius of the inside of the pipe.

The question states that the difference between the outside surface area and the inside surface area is 88 sq. cm:

SO - SI = 88 ; substituting:

(2 * pi * OR * 14) - (2 * pi * IR * 14) = 88 ; factoring

(2 * pi * 14) (OR - IR) = 88 ;dividing both sides by 2*pi*14

(OR - IR) = 88/(2 * pi) = 88/(2 * 3.14159*14) = 1.00

So, the outside radius is 1 cm more than the inside radius.

It's not clear if the volume stated is the volume of the metal in the pipe or the volume of air inside the pipe, so I will solve it both ways:

If volume of 176cc is of the air inside, the formula for this volume is 14 * pi * IR *IR

176 = 14 * 3.14159 * IR * IR ; dividing both sides by 14 * 3.14159

176/(14 * 3.14159) = IR * IR ; doing the arithmetic

4 = IR * IR ; taking the square root of both sides

sqrt(4) = IR

substituting back into the first equation, the OR is 1cm more than the IR, so

If volume of 176cc is of the iron in the pipe, the formula for that volume is the difference between the volume of the outside of the pipe and the volume of the inside of the pipe, or

(14 * pi * OR * OR) - (14 * pi * IR * IR) = 176 ; factoring

(14 * pi) ((OR * OR) - (IR * IR)) = 176 ; dividing both sides by 14 * pi

((OR * OR) - (IR * IR)) = 176/(14 * 3.14159) = 4

but, since OR is 1 cm more than IR (from above), we can substitute OR = IR + 1

and OR * OR = (IR + 1) * (IR + 1) = (IR*IR) + 2*IR +1

So, ((OR * OR) - (IR * IR)) = 4 becomes

(IR*IR) + 2*IR +1 - (IR*IR) = 4 ; simplifying (IRsquared - IRsquared = 0)

2*IR + 1 = 4 ; subtract 1 from both sides

2*IR = 3 ; divide both sides by 2

IR = 3/2 ;

Jan 11, 2011 | Mathsoft StudyWorks! Mathematics Deluxe...

Feb 01, 2017 | Canon EOS Rebel 2000 35mm SLR Camera

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