2.2 - formulas example 1: using the given values, solve for the variable in each formula that was...

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2.2 - Formulas Example 1: Using the given values, solve for the variable in each formula that was not assigned a value. Check: 9, 63 d rt t d 63 9 r 63 9 9 9 r 7 r 7 r 63 9 r 63 9 7 63 63

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Page 1: 2.2 - Formulas Example 1: Using the given values, solve for the variable in each formula that was not assigned a value. Check:

2.2 - Formulas

Example 1:

9, 63d rt t d

63 9r

63 9

9 9

r

7 r

Using the given values, solve for the variable in each formula that was not assigned a value.

7r

Check:

63 9r

63 97

63 63

Page 2: 2.2 - Formulas Example 1: Using the given values, solve for the variable in each formula that was not assigned a value. Check:

2.2 - Formulas

Example 2: Volume of a Pyramid1

40, 83

V Bh V h

40 81

3B 1

40 83

3 3 B

120 8B

15 B

120 8

8 8

B

15B

LCD: 3

Page 3: 2.2 - Formulas Example 1: Using the given values, solve for the variable in each formula that was not assigned a value. Check:

2.2 - Formulas

Example 3: Solve for the requested variable.

1

22 2A bh

1

2A bh

Area of a Triangle – solve for b

2A bh

2A bh

h h

2Ab

h

LCD: 2

Page 4: 2.2 - Formulas Example 1: Using the given values, solve for the variable in each formula that was not assigned a value. Check:

2.2 - FormulasExample 4: Solve for the requested variable.

932

532 32F C 9

325

F C

Celsius to Fahrenheit – solve for C

932

5F C

5 32 9F C

5 32

9

FC

932

55 5F C

5 32 9

9 9

F C

or 532

9F C

LCD: 5

Page 5: 2.2 - Formulas Example 1: Using the given values, solve for the variable in each formula that was not assigned a value. Check:

2.2 - FormulasExample 6: Solve for the requested variable.

Solve for v

h=π‘£π‘‘βˆ’ 16 𝑑2

+16 𝑑 2+16 𝑑2

h+16 𝑑2=𝑣𝑑

h+16 𝑑 2

𝑑=𝑣

h+16 𝑑 2

𝑑=𝑣𝑑𝑑

Page 6: 2.2 - Formulas Example 1: Using the given values, solve for the variable in each formula that was not assigned a value. Check:

2.2 - FormulasExample 5: Solve for the requested variable.

Solve for x

π‘Žπ‘₯βˆ’5=𝑐π‘₯βˆ’2βˆ’π‘π‘₯βˆ’π‘π‘₯π‘Žπ‘₯βˆ’π‘π‘₯βˆ’5=βˆ’2

+5+5π‘Žπ‘₯βˆ’π‘π‘₯=3π‘₯ (π‘Žβˆ’π‘)=3

π‘₯=3

π‘Žβˆ’π‘

π‘₯ (π‘Žβˆ’π‘)π‘Žβˆ’π‘

= 3π‘Žβˆ’π‘

Page 7: 2.2 - Formulas Example 1: Using the given values, solve for the variable in each formula that was not assigned a value. Check:

2.3 - ApplicationsSimple Interest

Simple Interest .Principal= Interest Rate

𝑰=𝑷𝑹𝑻

Interest Rate is stated as a percent and converted to a decimal for calculation purposes.

. Time

Time is stated in years or part of a year.

Page 8: 2.2 - Formulas Example 1: Using the given values, solve for the variable in each formula that was not assigned a value. Check:

61.25

Simple Interest

7%

Find the simple interest on a five year loan of $875 at a rate of 7%.

.$875

875 . 0.07

$306.25

. 5

. 5

. 5

306.25

𝑰=𝑷𝑹𝑻2.3 - Applications

Page 9: 2.2 - Formulas Example 1: Using the given values, solve for the variable in each formula that was not assigned a value. Check:

You invest $8000 in two accounts and earn a total of $323 in interest from both accounts in one year. The interest rates on the accounts were 4.6% and 2.8%. How much was invested in each account?

Total Interest = Account 1 + Account 2

𝑰=𝑷𝑹𝑻

Account 1

Account 2

Principal Interest Rate Interest

x

8000-x

πŸ’ .πŸ”%=𝟎 .πŸŽπŸ’πŸ”

𝟐 .πŸ–%=𝟎 .πŸŽπŸπŸ–

𝟎 .πŸŽπŸ’πŸ”π’™

𝟎 .πŸŽπŸπŸ– (πŸ–πŸŽπŸŽπŸŽβˆ’ 𝒙)

πŸ‘πŸπŸ‘=𝟎 .πŸŽπŸ’πŸ”π’™+𝟎 .πŸŽπŸπŸ– (πŸ–πŸŽπŸŽπŸŽβˆ’ 𝒙)πŸ‘πŸπŸ‘=𝟎 .πŸŽπŸ’πŸ”π’™+πŸπŸπŸ’βˆ’πŸŽ .πŸŽπŸπŸ–π’™πŸ—πŸ—=𝟎 .πŸŽπŸπŸ–π’™πŸ“πŸ“πŸŽπŸŽ=𝒙

Account 1: $5500

Account 2: 8000 – 5500

2.3 - Applications

Account 2: $2500

Page 10: 2.2 - Formulas Example 1: Using the given values, solve for the variable in each formula that was not assigned a value. Check:

x + 7

π’‚πŸ+π’ƒπŸ=π’„πŸx

2.3 - ApplicationsIn a right triangle, the length of the longer leg is 7 more inches than the shorter leg. The length of the hypotenuse is 8 more inches than the length of the shorter leg. Find the length of all three sides.

x + 8

(𝒙)𝟐+(𝒙+πŸ•)𝟐=(𝒙+πŸ–)𝟐

π’™πŸ+π’™πŸ+πŸπŸ’π’™+πŸ’πŸ—=π’™πŸ+πŸπŸ”π’™+πŸ”πŸ’βˆ’π’™πŸβˆ’πŸπŸ”π’™βˆ’πŸ”πŸ’βˆ’ π’™πŸβˆ’πŸπŸ”π’™βˆ’πŸ”πŸ’π’™πŸβˆ’πŸ π’™βˆ’πŸπŸ“=𝟎(𝒙+πŸ‘)(π’™βˆ’πŸ“)ΒΏπŸŽπ’™+πŸ‘=𝟎 π’™βˆ’πŸ“=πŸŽπ’™=βˆ’πŸ‘π’™=πŸ“

𝒙=πŸ“ π’Šπ’π’„π’‰π’†π’”π’™+πŸ•=πŸπŸπ’Šπ’π’„π’‰π’†π’”π’™+πŸ–=πŸπŸ‘π’Šπ’π’„π’‰π’†π’”

Page 11: 2.2 - Formulas Example 1: Using the given values, solve for the variable in each formula that was not assigned a value. Check:

π‘Ίπ’†π’π’π’Šπ’π’ˆπ’‘π’“π’Šπ’„π’†Γ—π’…π’π’˜π’π’‘π’‚π’šπ’Žπ’†π’π’•π’‘π’†π’“π’„π’†π’π’•π’‚π’ˆπ’†=π’…π’π’˜π’π’‘π’‚π’šπ’Žπ’†π’π’•

2.3 - ApplicationsA family paid $26,250 as a down payment for a home. This represents 15% of the selling price. What is the price of the home?

π’‘Γ—πŸŽ .πŸπŸ“ΒΏπŸπŸ”πŸπŸ“πŸŽπŸŽ .πŸπŸ“π’‘=πŸπŸ”πŸπŸ“πŸŽπŸŽ .πŸπŸ“π’‘πŸŽ .πŸπŸ“

=πŸπŸ”πŸπŸ“πŸŽπŸŽ .πŸπŸ“

𝒑=πŸπŸ•πŸ“πŸŽπŸŽπŸŽπ’‘=$πŸπŸ•πŸ“ ,𝟎𝟎𝟎

Page 12: 2.2 - Formulas Example 1: Using the given values, solve for the variable in each formula that was not assigned a value. Check:

2.3 - Applications

π’™πŸ“ 𝒙+πŸ”

𝒙+πŸ“ 𝒙+πŸ”

Special Pairs of Angles

Complimentary angles: Two angles whose sum is 90Β°. They are compliments of each other.

Supplementary angles: Two angles whose sum is 180Β°. They are supplements of each other.

One angle is six more than five times the other angle. What are their measurements if the are supplements of each other?

ΒΏπŸπŸ–πŸŽπŸ” 𝒙+πŸ”=πŸπŸ–πŸŽπŸ” 𝒙=πŸπŸ•πŸ’πŸ”π’™πŸ”

=πŸπŸ•πŸ’πŸ”

𝒙=πŸπŸ—

𝒙=πŸπŸ—Β°πŸ“ (πŸπŸ—)+πŸ”=πŸπŸ“πŸΒ°

πŸπŸ—Β°+πŸπŸ“πŸΒ°=πŸπŸ–πŸŽΒ°

Page 13: 2.2 - Formulas Example 1: Using the given values, solve for the variable in each formula that was not assigned a value. Check:

2.3 - ApplicationsA flower bed is in the shape of a triangle with one side twice the length of the shortest side, and the third side is 30 feet more than the length of the shortest side. Find the dimensions if the perimeter is 102 feet.

x = the length of the shortest side

2x = the length of the second side

x + 30 = the length of the third side

x 2x

x + 30

P = a + b + c

102 = x + 2x + x + 30

102 = 4x + 30

102 – 30 = 4x + 30 – 30

72 = 4x

4

4

4

72 x

π‘₯=18 𝑓𝑒𝑒𝑑2(18)=36 𝑓𝑒𝑒𝑑

18+30=48 𝑓𝑒𝑒𝑑

β†’ π‘₯=18

Page 14: 2.2 - Formulas Example 1: Using the given values, solve for the variable in each formula that was not assigned a value. Check:

2.3 - ApplicationsThe length of a rectangle is 4 less than twice the width. The area of the rectangle is 70. Find the dimensions of the rectangle.

h𝑙𝑒𝑛𝑔𝑑 =2 π‘₯βˆ’ 4

𝐴=𝑙×𝑀70=(2 π‘₯βˆ’ 4 ) π‘₯70=2 π‘₯2 βˆ’ 4 π‘₯0=2 π‘₯2βˆ’ 4 π‘₯βˆ’ 700=2 (π‘₯2 βˆ’2 π‘₯βˆ’35)0=2 (π‘₯+5)(π‘₯βˆ’ 7)

(π‘₯+5 )=0 (π‘₯βˆ’7 )=0π‘₯=βˆ’5 π‘₯=7

h𝑙𝑒𝑛𝑔𝑑 =2 π‘₯βˆ’ 4=2 (7 )βˆ’ 4=10

Page 15: 2.2 - Formulas Example 1: Using the given values, solve for the variable in each formula that was not assigned a value. Check:

2.3 - Applications

35 mph

π’…π’Šπ’”π’•π’‚π’π’„π’†=𝒓𝒂𝒕𝒆× π’•π’Šπ’Žπ’†40 mph

Two cars leave an airport at the same time. One is traveling due north at a rate of 40 miles per hour and the other is travelling due east at a rate of 35 miles per hour. When will the distance between the two cars be 110 miles?

110 mi.𝒅𝒏=πŸ’πŸŽπ’•π’‚πŸ+π’ƒπŸ=π’„πŸ

(πŸ’πŸŽπ’• )𝟐+(πŸ‘πŸ“π’• )𝟐=𝟏𝟏𝟎𝟐

πŸπŸ–πŸπŸ“π’•πŸ

πŸπŸ–πŸπŸ“=𝟏𝟐𝟏𝟎𝟎

πŸπŸ–πŸπŸ“

𝒅𝒆=πŸ‘πŸ“π’•

πŸπŸ”πŸŽπŸŽ π’•πŸ+πŸπŸπŸπŸ“π’•πŸ=πŸπŸπŸπŸŽπŸŽπŸπŸ–πŸπŸ“ π’•πŸ=𝟏𝟐𝟏𝟎𝟎

π’•πŸ=πŸ’ .πŸπŸ–πŸ‘ 𝒕=𝟐 .πŸŽπŸ•π’‰π’“π’” .β†’

Page 16: 2.2 - Formulas Example 1: Using the given values, solve for the variable in each formula that was not assigned a value. Check:

2.2 - Formulas

It takes Karen 3 hours to row a boat 30 kilometers upstream in a river. If the current was 4 kilometers per hour, how fast would she row in still water?

Rate Equation:

Rate upstream: (𝑑=π‘Ÿπ‘‘ ) 30=π‘Ÿπ‘’ (3) π‘Ÿπ‘’=10 hπ‘˜π‘

Rate in still water: π‘Ÿ 𝑠=10 hπ‘˜π‘ +4 hπ‘˜π‘ π‘Ÿ 𝑠=14 hπ‘˜π‘

How long would it take her to row 30 kilometers in still water? (𝑑=π‘Ÿπ‘‘ ) 30=14 𝑑 𝑑=2.14 hπ‘Ÿπ‘  .How long would it take her to row 30 kilometers downstream? (𝑑=π‘Ÿπ‘‘ ) 30=(14+4)𝑑 𝑑=1.67hπ‘Ÿπ‘  .30=18 𝑑

Page 17: 2.2 - Formulas Example 1: Using the given values, solve for the variable in each formula that was not assigned a value. Check:

2.3 - Applications