How do I solve this?

Step-by-step explanation:

From the picture attached,

Addition of the blocks in first row is 60

a + a + a + 12 = 60

3a + 12 = 60

3a = 60 - 12

3a = 48

a = 16

For second row,

(b + 5) + (b + 5) + (b + 5) + (b + 5) = 60

4(b + 5) = 60

b + 5 = 15

b = 10

For third row,

(a + b) + c = (b + 5) + (b + 5) + (b + 5)

a + b + c = 3(b + 5)

16 + 10 + c = 3(10 + 5) [Since, a = 16 and b = 10}

26 + c = 45

c = 45 - 26

c = 19

For fourth row,

3c + d = 60

3(19) + d = 60

57 + d = 60

d = 60 - 57

d = 3

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation: