Mr. Johnson has $15,000 to invest. Part of it is put in the bank at 4 percent, and part he puts in a savings and loan at 7 percent. If his yearly (simple) is $951, how much did he invest at each rate?

$3300 at 4%

$11700 at 7%

Step-by-step explanation:

Total investment = $15000

Total simple interest on investment = $951

Investment A:

Rate = 4% =0.04

Investment B:

Rate = 7% =0.07

Simple interest (S. I) = principal * rate * time

(S. I on investment A + S. I on investment B)

Let principal amount in investment A = A

principal amount in investment B = B

(A * 0.04 * 1) + (B * 0.07 * 1) = $951

0.04A + 0.07B = $951 - - - - (1)

A + B = $15000 - - - - (2)

from (2)

A = 15000 - B

Substitute A = 15000 - B into (1)

0.04(15000 - B) + 0.07B = 951

600 - 0.04B + 0.07B = 951

600 + 0.03B = 951

0.03B = 951 - 600

0.03B = 351

B = 351 / 0.03

B = $11700

From (2)

A + B = $15000

A + $11700 = $15000

A = $(15000 - 11700)

A = $3300

he used the algebra tiles correctly.

step-by-step explanation:

given : 3x(2x-1)

solution :

since he used algebra tiles .

armando's solution is the same in the solution shown in figure

as he distributed 2x into x and x and wrote them vertically

and also write   -1 vertically

and 3x into x and x and x and wrote it horizontally

performing the multiplication in the algebra tiles we get

and solving the given expression :

3x(2x-1)

since solving the expression using tiles   and solving the expression algebraically we get the same answer

thus   we can choose the first statement as the answer to this problem which is "he used the algebra tiles correctly".

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8