Brandon has 36000 to invest and wants to earn 5% interest per year. He will put some of the money into an account that earns 4.4% interest per year and the rest into an account that earns 6% interest per year. How much money should he put into each account so that he earns 5% annually

The money in an account that earns 4.4% interest per year = 22500

The rest into an account that earns 6% interest per year = 13500

Step-by-step explanation:

Interest = Principal × Rate × Time

Let x = money saved in account A

Let y = money saved in account B

Brandon has 36000 to invest and wants to earn 5% interest per year

x + y = 36000 Equation 1

x = 36000 - y

He will put some of the money into an account that earns 4.4% interest per year and the rest into an account that earns 6% interest per year.

Hence:

4.4% × x × 1 + 6% × y × 1 = 36000 × 5% × 1

0.044x + 0.06y= 1800 Equation 2

Substitute 36000-y for x in Equation 2

0.044(36000 - y) + 0.06y = 1800

1584 -0.044y + 0.06y = 1800

-0.044y + 0.06y = 1800 - 1584

0.016y = 216

y = 216/0.016

y = 13500

Note that:

x = 36000 - y

x = 36000 - 13500

x = 2250

Therefore,

The money in an account that earns 4.4% interest per year = 22500

The rest into an account that earns 6% interest per year = 13500

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step-by-step explanation:

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: