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stich3 [128]
3 years ago
7

Complete the formulas for surface area and volume of a cube, where a is the length of an edge

Mathematics
1 answer:
Amanda [17]3 years ago
8 0
Idk .....&;&&3&&26:)8/&9/&/!;)6,)3!299//!/!$:
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A chocolate maker melts three blocks of chocolate weighing 5.4 ounces, 6.1 ounces, and 5.7 ounces. If
Nady [450]

Answer:

43 truffles

Step-by-step explanation:

                       First add all of the weights together

5.4+6.1+5.7=17.2

                      Divide the total chocolate by the weight of a truffle

17.2/0.4=43

4 0
3 years ago
Bir satıcı sattığı her üründen satış fiyatının 2/9'u kadar kar etmektedir. Bu satıcı 1800 liraya sattığı bir televizyondan kaç l
drek231 [11]

Answer:

400 lira

Step-by-step explanation:

Yukarıdaki sorudan şunu söyleriz: Bir satıcı, sattığı her ürün için satış fiyatının 2 / 9'u kadar kar eder.

Dolayısıyla 1800 liraya televizyon satarsa ​​o televizyondan elde ettiği kâr şöyle hesaplanır:

2/9 × 1800 lira

= 400 lira

6 0
2 years ago
Factor out the coefficient of the variable 1/3b-1/3
Darya [45]
The variable is b, the coefinet is 1/3

remember
ab-ac=a(b-c)
1/3b-1/3=1/3(b-1)
there, factored
8 0
3 years ago
In the year 2005, a total of 750 fish were introduced into a manmade lake. The fish population was expected to grow at a rate of
rewona [7]
Exponential change can be modeled by:
N = N(0) x λⁿ; where λ is the rate of change. It is greater than 1 when there is growth and less than 1 when there is decay. n is the number of time periods passed.
λ is 1 + 0.083 = 1.083
N(0) is the initial value of 750
n is 2050 - 2005 = 45 years
N = 750 x (1.083)⁴⁵
= 27,000 fish
6 0
3 years ago
Read 2 more answers
) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra
artcher [175]

Answer:

y(t)=2e^{3t}(2-5t)

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2  

Using the theorem of the Laplace transform for derivatives, we know that:

\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)

Replacing the initial values y(0)=4, y′(0)=2 we obtain

\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4

and our differential equation (*) gets transformed in the algebraic equation

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0

Solving for Y(s) we get

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}

Now, we brake down the rational expression of Y(s) into partial fractions

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here  

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}

and

\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}

we know that

\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}

and for the first translation property of the inverse Laplace transform

\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}

and the solution of our differential equation is

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}

5 0
3 years ago
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