Hello,
8+h>2+3h
==>-2h>2-8
==>-2h>-6
==>h<3
Answer C
Answer: Parallel: y=-2/7x-2 1/7 Perpendicular: y=7/2x+13
Step-by-step explanation:
2x+7y=14
Subtract 2x from both sides.
7y=-2x+14
Divide both sides by 7 to isolate y.
y=-2/7x+2
Parallel:
Plug in x and y with same slope as the original.
-1=-2/7(-4)+b
Solve for b:
-1=8/7+b
-2 1/7=b
y=-2/7x-2 1/7
Perpendicular:
Plug in x and y with the negative inverse of the original slope.
-1=7/2(-4)+b
Solve for b.
-1=-28/2+b
-1=-14+b
13=b
y=7/2x+13
Since it’s asking for square inches on a rectangular prism, this is solving for surface area. So we use the following formula:
2(h*w)+2(h*l)+2(w*l)=A
2(10*8)+2(10*12)+2(8*12)
2(80)+2(120)+2(96)
160+240+192
592
592 sq. in.
Width = x
length = x+10
Area of deck = 144 square feet
Area = length * width
144 = x(x+10)
Solving for x in a quadratic:
x²+10x = 144
x²+10x-144 = 0
Factor:
(x+18)(x-8) = 0
Solving for x:
x = 8, x = -18
Dimensions cannot be negative, therefore x = 8, only.
Length = x + 10
Length = 18
Width = 8
Answer: The required solution is
Step-by-step explanation: We are given to solve the following differential equation :
Let us consider that
be an auxiliary solution of equation (i).
Then, we have
Substituting these values in equation (i), we get
![m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.](https://tex.z-dn.net/?f=m%5E2e%5E%7Bmt%7D%2B10me%5E%7Bmt%7D%2B25e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%5E2%2B10y%2B25%29e%5E%7Bmt%7D%3D0%5C%5C%5C%5C%5CRightarrow%20m%5E2%2B10m%2B25%3D0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7Bsince%20%7De%5E%7Bmt%7D%5Cneq0%5D%5C%5C%5C%5C%5CRightarrow%20m%5E2%2B2%5Ctimes%20m%5Ctimes5%2B5%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%2B5%29%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20m%3D-5%2C-5.)
So, the general solution of the given equation is
Differentiating with respect to t, we get
According to the given conditions, we have
and
Thus, the required solution is
