I’m not sure which of these you meant to type so I answered both ways
B=(RT-4)/6 multiply by 6
6B= RT-4. Add 4
6B+4= RT. Divide by R
(6B+4)/R = T
Or B=RT -4/6. Multiply by 6
6B=6RT -4. Add 4
6B+4= 6RT. Divide by 6R
(6B-4)/6R = T. Reduce
(3B-2)/3R = T
Answer:
t = 4 seconds
Step-by-step explanation:
The height of the projectile after it is launched is given by the function :

t is time in seconds
We need to find after how many seconds will the projectile land back on the ground. When it land, h(t)=0
So,

The above is a quadratic equation. It can be solved by the formula as follows :

Here, a = -16, b = 32 and c = 128

Neglecting negative value, the projectile will land after 4 seconds.
M<A = <span>20°
m<B = m<C = 8</span><span>0°
</span>Law of Sines , in any triangle we have
a/sin A = b/sin B = c/sin c<span>
4/sin20 = AC/sin80 = AB/sin80
now we can solve AC
</span>4/sin20 = AC/sin80
<span>AC = 4 (sin80)/ sin20
AC = 4(0.98) / (0.34)
AC = 3.92 / 0.34
AC = 11.52
answer
</span>C.11.52 centimeters<span>
</span>
3/4x-12-5=15
3/4x-17=15
3x-68=60
3x=60+68
3x=128
X=128/3
Answer:
11 and 12
Step-by-step explanation:
x(x+1)=132
x^2+x=132
x^2+x-132=0
(x-11)(x+12)=0
x = 11 or x = -12
when x=11 ,
x+1=12
11*12=132