Side JL is 4√2 recall that in a 30-60-90 right triangle the hypotenuse is 2 times the size of the short leg.
JL also serves as the hypotenuse of the 45-45-90 triangle JML. The ratio of side lengths in this triangle is 1:1:√2
So we can see that the value of x = 4
Answer: E - S = (-16 and 6)
Step-by-step explanation:1/3 of the 30 decimals in T have an even tenths digit, it follows that 1/3*(30)=10 decimals in T have an even tenths digit.
Hence: Te =list of 10 decimals
Se = sum of all 10 decimals in Te
Ee =estimated sum of all 10 decimals in Te after rounding up.
Remaining 20 decimals in T all have an odd tenths digits.
To =list of this 20 decimals
So = sum of all 20 decimals in To
Eo = estimated sum of 20 decimals in To
Hence,
E = Ee + Eo and S =Se +So, hence:
E-S, =(Ee+Eo) - (Se+So) =(Ee-Se) +(Eo-So)
Ee-Se >10 (0.1)=1
S=10(1.8)+20(1.9) =18+38=56
E=10(2)+20(1)=40
E-S =40-56=-16.
AlsoS=10(1.2)+20(1.1)=34
E=10(2)+20(1)=40
E-S=40-34=6
Answer:
yp = -x/8
Step-by-step explanation:
Given the differential equation: y′′−8y′=7x+1,
The solution of the DE will be the sum of the complementary solution (yc) and the particular integral (yp)
First we will calculate the complimentary solution by solving the homogenous part of the DE first i.e by equating the DE to zero and solving to have;
y′′−8y′=0
The auxiliary equation will give us;
m²-8m = 0
m(m-8) = 0
m = 0 and m-8 = 0
m1 = 0 and m2 = 8
Since the value of the roots are real and different, the complementary solution (yc) will give us
yc = Ae^m1x + Be^m2x
yc = Ae^0+Be^8x
yc = A+Be^8x
To get yp we will differentiate yc twice and substitute the answers into the original DE
yp = Ax+B (using the method of undetermined coefficients
y'p = A
y"p = 0
Substituting the differentials into the general DE to get the constants we have;
0-8A = 7x+1
Comparing coefficients
-8A = 1
A = -1/8
B = 0
yp = -1/8x+0
yp = -x/8 (particular integral)
y = yc+yp
y = A+Be^8x-x/8
Answer:
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Answer:
x = -1
Step-by-step explanation:
By graphing x = -1, you will see that the line splits this rhombus in two equal halves
