The value of h(t) when
is 10.02.
Solution:
Given function 
To find the value of h(t) when
:

Substitute
in the given function.


Now multiply the common terms into inside the bracket.

Now, in the first term, the numerator and denominator both have common factor 16. So reduce the first term into the lowest term.

To make the denominator same, take LCM of the denominators.
LCM of 64 and 32 = 64




= 10.02

Hence the value of h(t) when
is 10.02.
1.) y+6=3(x+2) C
y+6=3x+6
y=3x+6-6
y=3x+6
2.) y=1/2(x+8)-2 B
y=1/2x+4-2
y=1/2x+2
3.) y+1=1(x-3) E
y+1=x-3
y=x-3-1
y=x-4
4.) -4x+y=-2 A
y=-4x-2
5.) 2x-4y=-4 F
-4y=-2x-4
y=-2/-4x-4/-4
y=2/4x+4/4
y=1/2x+1
6.) 2x+4y=8 D
4y=-2x+8
y=-2/4x+8/4
y=-1/2x+4/2
y=-1/2x+2
-2x + xy = 30.....when y = 8
-2x + 8x = 30
6x = 30
x = 30/6
x = 5 <==
Your answe would be 1 9/13
Are those supposed to be exponents? When multiplying exponents you add them together
b^3 * b^1 * b^4 * b^2
3 + 1 + 4 + 2 = 10
Answer: b^10