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wariber [46]
3 years ago
6

A block of sugar maple wood has a mass of 82.8 g and density of 0.69 ​ g/cm3 ​. The block is a rectangular prism with a length o

f 5 cm and a width of 4 cm.
What is the height of the block of sugar maple?
Mathematics
1 answer:
Dima020 [189]3 years ago
8 0

For anyone seeing this that is in k12 the answer is definitely 6. Don't listen to the guy saying it's 9. Have a nice day :)


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−15 ÷ 5 = A number line going from negative 30 to positive 30. An arrow goes from 5 to 10, from 10 to 15, from 15 to 20.
miv72 [106K]

Answer:

-3

Step-by-step explanation:

-15 divided by 5= -3

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6) If the perimeter of the equilateral triangle ABC is x inches. The mid points of the sides are joined to form the smaller tria
Nataliya [291]

Answer:

96 in

Step-by-step explanation:

If the midpoints of the sides are joined to form the smaller triangle, then the perimeter of the smaller triangle is half the perimeter of the greater triangle, because the sides of the smaller triangle are midlines of the greater triangle. By the triangle's midline theorem, each triangle's midline is half the side to which this midline is parallel.

So, if the perimeter of 6th triangle is 3 inches, then the perimeter of 5th triangle is 6 inches, the perimeter of 4th triangle is 12 inches, the perimeter of 3rd triangle is 24 in, the perimeter of 2nd triangle is 48 in and the perimeter of the initial triangle is 96 in

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3 years ago
Solve the following differential equation using using characteristic equation using Laplace Transform i. ii y" +y sin 2t, y(0) 2
kifflom [539]

Answer:

The solution of the differential equation is y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)

Step-by-step explanation:

The differential equation is given by: y" + y = Sin(2t)

<u>i) Using characteristic equation:</u>

The characteristic equation method assumes that y(t)=e^{rt}, where "r" is a constant.

We find the solution of the homogeneus differential equation:

y" + y = 0

y'=re^{rt}

y"=r^{2}e^{rt}

r^{2}e^{rt}+e^{rt}=0

(r^{2}+1)e^{rt}=0

As e^{rt} could never be zero, the term (r²+1) must be zero:

(r²+1)=0

r=±i

The solution of the homogeneus differential equation is:

y(t)_{h}=c_{1}e^{it}+c_{2}e^{-it}

Using Euler's formula:

y(t)_{h}=c_{1}[Sin(t)+iCos(t)]+c_{2}[Sin(t)-iCos(t)]

y(t)_{h}=(c_{1}+c_{2})Sin(t)+(c_{1}-c_{2})iCos(t)

y(t)_{h}=C_{1}Sin(t)+C_{2}Cos(t)

The particular solution of the differential equation is given by:

y(t)_{p}=ASin(2t)+BCos(2t)

y'(t)_{p}=2ACos(2t)-2BSin(2t)

y''(t)_{p}=-4ASin(2t)-4BCos(2t)

So we use these derivatives in the differential equation:

-4ASin(2t)-4BCos(2t)+ASin(2t)+BCos(2t)=Sin(2t)

-3ASin(2t)-3BCos(2t)=Sin(2t)

As there is not a term for Cos(2t), B is equal to 0.

So the value A=-1/3

The solution is the sum of the particular function and the homogeneous function:

y(t)= - \frac{1}{3} Sin(2t) + C_{1} Sin(t) + C_{2} Cos(t)

Using the initial conditions we can check that C1=5/3 and C2=2

<u>ii) Using Laplace Transform:</u>

To solve the differential equation we use the Laplace transformation in both members:

ℒ[y" + y]=ℒ[Sin(2t)]

ℒ[y"]+ℒ[y]=ℒ[Sin(2t)]  

By using the Table of Laplace Transform we get:

ℒ[y"]=s²·ℒ[y]-s·y(0)-y'(0)=s²·Y(s) -2s-1

ℒ[y]=Y(s)

ℒ[Sin(2t)]=\frac{2}{(s^{2}+4)}

We replace the previous data in the equation:

s²·Y(s) -2s-1+Y(s) =\frac{2}{(s^{2}+4)}

(s²+1)·Y(s)-2s-1=\frac{2}{(s^{2}+4)}

(s²+1)·Y(s)=\frac{2}{(s^{2}+4)}+2s+1=\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)}

Y(s)=\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)(s^{2}+1)}

Y(s)=\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}

Using partial franction method:

\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}=\frac{As+B}{s^{2}+4} +\frac{Cs+D}{s^{2}+1}

2s^{3}+s^{2}+8s+6=(As+B)(s²+1)+(Cs+D)(s²+4)

2s^{3}+s^{2}+8s+6=s³(A+C)+s²(B+D)+s(A+4C)+(B+4D)

We solve the equation system:

A+C=2

B+D=1

A+4C=8

B+4D=6

The solutions are:

A=0 ; B= -2/3 ; C=2 ; D=5/3

So,

Y(s)=\frac{-\frac{2}{3} }{s^{2}+4} +\frac{2s+\frac{5}{3} }{s^{2}+1}

Y(s)=-\frac{1}{3} \frac{2}{s^{2}+4} +2\frac{s }{s^{2}+1}+\frac{5}{3}\frac{1}{s^{2}+1}

By using the inverse of the Laplace transform:

ℒ⁻¹[Y(s)]=ℒ⁻¹[-\frac{1}{3} \frac{2}{s^{2}+4}]-ℒ⁻¹[2\frac{s }{s^{2}+1}]+ℒ⁻¹[\frac{5}{3}\frac{1}{s^{2}+1}]

y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)

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3 years ago
Two lines that have the same slopes, when will they intersect
taurus [48]

Answer: No. If two lines have the same slope they are parallel lines (or they are the same line), so they cannot intersect at one point. Because the triangles are congruent, the angles where the lines meet the x-axis are congruent. Because the lines form the same angle with the x-axis, they are parallel.

Step-by-step explanation:

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1. Jan spends part of her year as a member of a
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Answer:

13 months in a half......

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