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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

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

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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=𝑣𝑡− 16 𝑡2

+16 𝑡 2+16 𝑡2

h+16 𝑡2=𝑣𝑡

h+16 𝑡 2

𝑡=𝑣

h+16 𝑡 2

𝑡=𝑣𝑡𝑡

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

𝑎𝑥−5=𝑐𝑥−2−𝑐𝑥−𝑐𝑥𝑎𝑥−𝑐𝑥−5=−2

+5+5𝑎𝑥−𝑐𝑥=3𝑥 (𝑎−𝑐)=3

𝑥=3

𝑎−𝑐

𝑥 (𝑎−𝑐)𝑎−𝑐

= 3𝑎−𝑐

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

𝟒 .𝟔%=𝟎 .𝟎𝟒𝟔

𝟐 .𝟖%=𝟎 .𝟎𝟐𝟖

𝟎 .𝟎𝟒𝟔𝒙

𝟎 .𝟎𝟐𝟖 (𝟖𝟎𝟎𝟎− 𝒙)

𝟑𝟐𝟑=𝟎 .𝟎𝟒𝟔𝒙+𝟎 .𝟎𝟐𝟖 (𝟖𝟎𝟎𝟎− 𝒙)𝟑𝟐𝟑=𝟎 .𝟎𝟒𝟔𝒙+𝟐𝟐𝟒−𝟎 .𝟎𝟐𝟖𝒙𝟗𝟗=𝟎 .𝟎𝟏𝟖𝒙𝟓𝟓𝟎𝟎=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

𝒂𝟐+𝒃𝟐=𝒄𝟐x

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+𝟕)𝟐=(𝒙+𝟖)𝟐

𝒙𝟐+𝒙𝟐+𝟏𝟒𝒙+𝟒𝟗=𝒙𝟐+𝟏𝟔𝒙+𝟔𝟒−𝒙𝟐−𝟏𝟔𝒙−𝟔𝟒− 𝒙𝟐−𝟏𝟔𝒙−𝟔𝟒𝒙𝟐−𝟐 𝒙−𝟏𝟓=𝟎(𝒙+𝟑)(𝒙−𝟓)¿𝟎𝒙+𝟑=𝟎 𝒙−𝟓=𝟎𝒙=−𝟑𝒙=𝟓

𝒙=𝟓 𝒊𝒏𝒄𝒉𝒆𝒔𝒙+𝟕=𝟏𝟐𝒊𝒏𝒄𝒉𝒆𝒔𝒙+𝟖=𝟏𝟑𝒊𝒏𝒄𝒉𝒆𝒔

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:

𝑺𝒆𝒍𝒍𝒊𝒏𝒈𝒑𝒓𝒊𝒄𝒆×𝒅𝒐𝒘𝒏𝒑𝒂𝒚𝒎𝒆𝒏𝒕𝒑𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆=𝒅𝒐𝒘𝒏𝒑𝒂𝒚𝒎𝒆𝒏𝒕

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?

𝒑×𝟎 .𝟏𝟓¿𝟐𝟔𝟐𝟓𝟎𝟎 .𝟏𝟓𝒑=𝟐𝟔𝟐𝟓𝟎𝟎 .𝟏𝟓𝒑𝟎 .𝟏𝟓

=𝟐𝟔𝟐𝟓𝟎𝟎 .𝟏𝟓

𝒑=𝟏𝟕𝟓𝟎𝟎𝟎𝒑=$𝟏𝟕𝟓 ,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

𝒙𝟓 𝒙+𝟔

𝒙+𝟓 𝒙+𝟔

Special Pairs of Angles

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

Supplementary angles: Two angles whose sum is 180°. 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?

¿𝟏𝟖𝟎𝟔 𝒙+𝟔=𝟏𝟖𝟎𝟔 𝒙=𝟏𝟕𝟒𝟔𝒙𝟔

=𝟏𝟕𝟒𝟔

𝒙=𝟐𝟗

𝒙=𝟐𝟗°𝟓 (𝟐𝟗)+𝟔=𝟏𝟓𝟏°

𝟐𝟗°+𝟏𝟓𝟏°=𝟏𝟖𝟎°

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

𝑥=18 𝑓𝑒𝑒𝑡2(18)=36 𝑓𝑒𝑒𝑡

18+30=48 𝑓𝑒𝑒𝑡

→ 𝑥=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.

h𝑙𝑒𝑛𝑔𝑡 =2 𝑥− 4

𝐴=𝑙×𝑤70=(2 𝑥− 4 ) 𝑥70=2 𝑥2 − 4 𝑥0=2 𝑥2− 4 𝑥− 700=2 (𝑥2 −2 𝑥−35)0=2 (𝑥+5)(𝑥− 7)

(𝑥+5 )=0 (𝑥−7 )=0𝑥=−5 𝑥=7

h𝑙𝑒𝑛𝑔𝑡 =2 𝑥− 4=2 (7 )− 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

𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆=𝒓𝒂𝒕𝒆× 𝒕𝒊𝒎𝒆40 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=𝟒𝟎𝒕𝒂𝟐+𝒃𝟐=𝒄𝟐

(𝟒𝟎𝒕 )𝟐+(𝟑𝟓𝒕 )𝟐=𝟏𝟏𝟎𝟐

𝟐𝟖𝟐𝟓𝒕𝟐

𝟐𝟖𝟐𝟓=𝟏𝟐𝟏𝟎𝟎

𝟐𝟖𝟐𝟓

𝒅𝒆=𝟑𝟓𝒕

𝟏𝟔𝟎𝟎 𝒕𝟐+𝟏𝟐𝟐𝟓𝒕𝟐=𝟏𝟐𝟏𝟎𝟎𝟐𝟖𝟐𝟓 𝒕𝟐=𝟏𝟐𝟏𝟎𝟎

𝒕𝟐=𝟒 .𝟐𝟖𝟑 𝒕=𝟐 .𝟎𝟕𝒉𝒓𝒔 .→

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=𝑟𝑡 ) 30=𝑟𝑢 (3) 𝑟𝑢=10 h𝑘𝑝

Rate in still water: 𝑟 𝑠=10 h𝑘𝑝 +4 h𝑘𝑝 𝑟 𝑠=14 h𝑘𝑝

How long would it take her to row 30 kilometers in still water? (𝑑=𝑟𝑡 ) 30=14 𝑡 𝑡=2.14 h𝑟𝑠 .How long would it take her to row 30 kilometers downstream? (𝑑=𝑟𝑡 ) 30=(14+4)𝑡 𝑡=1.67h𝑟𝑠 .30=18 𝑡

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