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Sedaia [141]
3 years ago
15

The sum of 4x² + x - 8 and x² + 9 can be expressed as​

Mathematics
2 answers:
kotykmax [81]3 years ago
5 0

Answer:

5x^2+x+1

Step-by-step explanation:

marta [7]3 years ago
3 0

<h3>{4x}^{2}  + x - 8 +  {x}^{2}  + 9 \\  {4x}^{2}  +  {x}^{2}  + x - 8 + 9 \\  {5x}^{2}  + x + 1</h3>

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If you multiply 235 by 49, what will be the partial products?
Olenka [21]

Answer:

Step-by-step explanation:

(200+30+5)(40+9)

40*200 = 8000

40*30=   1200

40*5=200

9*200= 1800

9*30=270

9*5=45

8000+1200+200+1800+270+45 =11515

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swat32

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it 2 parts away

Step-by-step explanation:

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The sum of the first 20 positive odd integers
svet-max [94.6K]

Answer:

400

Step-by-step explanation:

We\ know\ that,\\'The\ sum\ of\ n\ no.\ positive\ odd\ integers\ is\ given\ by\ the\ formula:\\S_O=n^2\\Hence,\\Here,\\As\ n=20,\\The\ sum\ of\ the\ first\ 20\ positive\ odd\ integers=20^2=400\\

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3 years ago
I need help, can anyone help me??? I'll give anyone brainliest for correct answers
elena55 [62]
I think it’s b the water is 2.2 meters below the ground floor. Sorry if I’m wrong.
4 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
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