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marysya [2.9K]
3 years ago
13

Someone help. (Find the area of a rectangle with a lenght of 5a^2 b^4 and a width of (3ab^3)^2

Mathematics
1 answer:
Dmitry_Shevchenko [17]3 years ago
3 0

Answer:

All you do is just multiply them.

Step-by-step explanation:

5a^2 b^4(3ab^3)^2=

45(a^(4))(b^(10))

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Simplify: 8.2c−0.9+(5.4−18d+7.2c)÷0.06
maxonik [38]
8.2c−0.9+(5.4−18d+7.2c)÷0.06   Distribute:<span>  =<span><span><span><span><span><span><span>8.2c</span>+</span>−0.9</span>+<span>120c</span></span>+</span>−<span>300d</span></span>+<span>90

</span></span></span>   <span>=<span><span><span>(<span><span>8.2c</span>+<span>120c</span></span>)</span>+<span>(<span>−<span>300d</span></span>)</span></span>+<span>(<span><span>−0.9</span>+90</span>)</span></span></span><span>   =<span><span><span>128.2c</span>+<span>−<span>300d</span></span></span>+89.1</span></span> Answer:<span>=<span><span><span>128.2c</span>−<span>300d</span></span>+<span>89.1</span></span></span>

5 0
3 years ago
You have driving lessons every third day in swimming lessons every fifth day today you have both classes how many days will you
hjlf
To find the answer, you need to find the LCM of 3 and 5. It's 15, meaning that every 15 days, you will have both lessons on the same day. Have an awesome day! :)
8 0
3 years ago
PLEASE HELP DUE SOON
marishachu [46]

Answer:

Step-by-step explanation:

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6 0
3 years ago
A company wishes to manufacture some boxes out of card. The boxes will have 6 sides (i.e. they covered at the top). They wish th
Serhud [2]

Answer:

The dimensions are, base b=\sqrt[3]{200}, depth d=\sqrt[3]{200} and height h=\sqrt[3]{200}.

Step-by-step explanation:

First we have to understand the problem, we have a box of unknown dimensions (base b, depth d and height h), and we want to optimize the used material in the box. We know the volume V we want, how we want to optimize the card used in the box we need to minimize the Area A of the box.

The equations are then, for Volume

V=200cm^3 = b.h.d

For Area

A=2.b.h+2.d.h+2.b.d

From the Volume equation we clear the variable b to get,

b=\frac{200}{d.h}

And we replace this value into the Area equation to get,

A=2.(\frac{200}{d.h} ).h+2.d.h+2.(\frac{200}{d.h} ).d

A=2.(\frac{200}{d} )+2.d.h+2.(\frac{200}{h} )

So, we have our function f(x,y)=A(d,h), which we have to minimize. We apply the first partial derivative and equalize to zero to know the optimum point of the function, getting

\frac{\partial A}{\partial d} =-\frac{400}{d^2}+2h=0

\frac{\partial A}{\partial h} =-\frac{400}{h^2}+2d=0

After solving the system of equations, we get that the optimum point value is d=\sqrt[3]{200} and  h=\sqrt[3]{200}, replacing this values into the equation of variable b we get b=\sqrt[3]{200}.

Now, we have to check with the hessian matrix if the value is a minimum,

The hessian matrix is defined as,

H=\left[\begin{array}{ccc}\frac{\partial^2 A}{\partial d^2} &\frac{\partial^2 A}{\partial d \partial h}\\\frac{\partial^2 A}{\partial h \partial d}&\frac{\partial^2 A}{\partial p^2}\end{array}\right]

we know that,

\frac{\partial^2 A}{\partial d^2}=\frac{\partial}{\partial d}(-\frac{400}{d^2}+2h )=\frac{800}{d^3}

\frac{\partial^2 A}{\partial h^2}=\frac{\partial}{\partial h}(-\frac{400}{h^2}+2d )=\frac{800}{h^3}

\frac{\partial^2 A}{\partial d \partial h}=\frac{\partial^2 A}{\partial h \partial d}=\frac{\partial}{\partial h}(-\frac{400}{d^2}+2h )=2

Then, our matrix is

H=\left[\begin{array}{ccc}4&2\\2&4\end{array}\right]

Now, we found the eigenvalues of the matrix as follow

det(H-\lambda I)=det(\left[\begin{array}{ccc}4-\lambda&2\\2&4-\lambda\end{array}\right] )=(4-\lambda)^2-4=0

Solving for\lambda, we get that the eigenvalues are:  \lambda_1=2 and \lambda_2=6, how both are positive the Hessian matrix is positive definite which means that the functionA(d,h) is minimum at that point.

4 0
3 years ago
Given points S(-4,-2) and T(-1,0) What is the length of ST? Round to the nearest tenth.
zheka24 [161]

Answer:

the length of ST = 3.6 to the nearest tenth

Step-by-step explanation:

Points on the Cartesian are written as (x, y)

Given points S(-4, -2) and T(-1, 0)

The distance between these two points is given by the formula

ST = √[(x₂-x₁)² + (y₂-y₁)²]

x₂ = -4

x₁ = -1

y₂ = -2

y₁ = 0

ST = √{[-4 - (-1)]² + [-2 - 0]²}

ST = √[(-4 + 1)² + (-2)²]

ST = √(-3²+4)

ST = √(9 + 4)

ST = √13

ST = 3.6 to the nearest tenth

6 0
3 years ago
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