10.
12*3^2+36 needs to be 540
12*9+36
9+36=45
12*45=540 so
12(3^2+36)
11.
32/2^3-4 needs to be 8
32/(2^3-4)
12.
2.3^2+9*4/2 needs to be 28.58
5.29+9*4/2
(5.29+9)*4/2=28.58
(2.3^2+9)*4/2
Hope this helps :)
If <em>f(x)</em> is a linear function, then
<em>f(x)</em> = <em>ax</em> + <em>b</em>
for some constants <em>a</em> and <em>b</em>.
Given that <em>f</em> (1) = 4 and <em>f</em> (4) = 1, these constants are such that
<em>f</em> (1) = <em>a</em> + <em>b</em> = 4
<em>f</em> (4) = 4<em>a</em> + <em>b</em> = 1
Solve for <em>a</em> and <em>b</em> : eliminate <em>b</em> by subtracting the two equations,
(<em>a</em> + <em>b</em>) - (4<em>a</em> + <em>b</em>) = 4 - 1
-3<em>a</em> = 3
<em>a</em> = -1
Then
-1 + <em>b</em> = 4
<em>b</em> = 5
So we have
<em>f(x)</em> = -<em>x</em> + 5
About 40 swimmers maybe more
a) 70.1 m
The ball is moving by uniformly accelerated motion, with constant acceleration 
(acceleration due to gravity) towards the ground. The height of the ball at time t is given by the equation
where 
is the height of the ball at time t=0. Substituting t=1 s, we can find the height of the ball 1 seconds after it has been dropped:
b) 3.9 s
We can still use the same equation we used in the previous part of the problem:
This time, we want to find the time t at which the ball hits the ground, which means the time t at which h(t)=0. So we have
And solving for t we find
