Answer:
-2/3x+6
Step-by-step explanation:
![\bf \stackrel{\textit{multiplying both sides by LCD of 3}}{3(y+5)=3\left[ \cfrac{5}{3}(x-3) \right]}\implies 3y+15=5(x-3) \\\\\\ 3y+15=5x-15\implies -5x+3y=-30\implies \stackrel{\textit{multiplying by -1}}{5x-3y=30}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20LCD%20of%203%7D%7D%7B3%28y%2B5%29%3D3%5Cleft%5B%20%5Ccfrac%7B5%7D%7B3%7D%28x-3%29%20%5Cright%5D%7D%5Cimplies%203y%2B15%3D5%28x-3%29%0A%5C%5C%5C%5C%5C%5C%0A3y%2B15%3D5x-15%5Cimplies%20-5x%2B3y%3D-30%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20by%20-1%7D%7D%7B5x-3y%3D30%7D)
bearing in mind the standard form uses all integers, and the x-variable cannot have a negative coefficient.
Surface of a cubical box=6(side²)
1)We have to calculate the surface of this cubical box.
Rate=cost of painting / surface ⇒surface=cost of painting/rate
Data:
Rate=$15/m²
cost of painting=$1440
Surface=$1440/($15/m²)=96 m²
2)We find out the length of the side:
Surface of a cubical box=6(side²)
Data:
Surface of a cubical box=96 m2
Therefore:
96m²=6 (side²)
side²=96 m²/6
side²=16 m²
side=√(16 m²)=4 m
3) We find the volume of a cubical box.
volume=(side³)
volume=(4 m)³
volume=64 m³
Answer: the volume of this cubical box would be 64 m³.
Supplementary angles sum up to 180°, suppose the supplement of 161° is x. Then
161+x=180
solving for x we subtract 161 from both sides
161-161+x=180-161
x=19
thus the supplement is 19°
1.
the x value of the vertex in form
ax^2+bx+c=y
is
-b/2a
so
-2x^2+8x-18
x value of vertex is
-8/(2*-2)=-8/-4=2
plug it in to get y value
-2(2)^2+8(2)-18
-2(4)+16-18
-8-2
-10
vertex is at (2,-10)
or you could complete the square to get into y=a(x-h)^2+k, where the vertex is (h,k)
so as follows
y=(-2x^2+8x)-18
y=-2(x^2-4x)-18
y=-2(x^2-4x+4-4)-18
y=-2((x-2)^2-4)-18
y=-2(x-2)^2+8-18
y=-2(x-2)^2-10
vertex is (2,-10)
5.
vertex is the time where the speed is the highest
at about t=10, the speed is at its max
