Answer: 3f - 11 = painted turtles
Step-by-step explanation
3 times the bullfrogs - 3f
Subtract 11 - 3f - 11
Given:
The graph of linear function.
To find:
The slope intercept form of the linear equation from the given graph.
Solution:
The slope intercept form of a linear function is:
From the given graph it is clear that the graph passes through the points (1,300) and (4,800).
The equation of line is
Adding 300 on both sides, we get
Therefore, the required equation is .
Answer:
225π
Step-by-step explanation:
Since we are given that m is equivalent to 15 we can plug it into the Area equation given to us like so:
Area = πm²
Area = π15²
Area = π225
Assuming that the angles are congruent, then y = 29.
4y - 8 = 79 + y
4y = 87 + y Add 8 to both sides
3y = 87 Subtract y from both sides
y = 29
Answer:
51-54: Simple Interest. Calculate the amount of money you will have in the following accounts after 5 years, assuming that you eam simple interest 51. You deposit $ 700 in an account with an annual interest rate of 4% 52. You deposit $1200 in an account with an annual interest rate of 3% 53. You deposit $3200 in an account with an annual interest rate of 3.5% 54. You deposit $1800 in an account with an annual interest rate of 3.8% 55-56: Simple versus Compound Interest. Complete the following tables, which show the performance of two investments over a 5-year period. Round all figures to the nearest dollar. 55 Suzanne deposits $3000 in an account that earns simple interest at an annual rate of 2.5%. Derek deposits $3000 in an account that earns compound interest at an annual rate of 2.5%. Suzanne's Suzanne's Derek's Annual | Derek's Year Annual Interest Balance Interest Balance rest formula to the stated pe 57-62: Compound Interest. Use the compound interest form compute the balance in the following accounts after the state riod of time, assuming interest is compounded annually. 57. $10,000 is invested at an APR of 4% for 10 years. 58. $10,000 is invested at an APR of 2.5% for 20 years. 59. $15,000 is invested at an APR of 3.2% for 25 years. 60. $3000 is invested at an APR of 1.8% for 12 years. 61. 55000 is invested at an APR of 3.1% for 12 years. 62. $ 40,000 is invested at an APR of 2.8% for 30 years. 63-70: Compounding More Than Once a Year. Use the appropriate compound interest formula to compute the balance in the following accounts after the stated period of time. 63. $10,000 is invested for 10 years with an APR of 2% and quarterly compounding. 64. $2000 is invested for 5 years with an APR of 3% and daily compounding 65. $25,000 is invested for 5 years with an APR of 3% and daily compounding 66. $10,000 is invested for 5 years with an APR of 2.75% and monthly compounding. 67. $2000 is invested for 15 years with an APR of 5% and monthly compounding 68. $30,000 is invested for 15 years with an APR of 4.5% ana daily compounding. 69. $25,000 is invested for 30 years with an APR of 3.7% quarterly compounding. 70. $15,000 is invested for 15 years with an APR of 4.2% monthly compounding. 71-74. Annual.
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