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3 years ago

14

What is the value of the expression given below?

2 answers:

natita [175]3 years ago

6 0

There is no expression so there is no value.

uranmaximum [27]3 years ago

6 0

Answer: I dont Know Search IT up.

Step-by-step explanation:

Hope This helps

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I WILL GIVE BRAINLIEST. PLEASE ANSWER!

Kazeer [188]

Answer:

Step-by-step explanation:

1 person = 240 granola per year

260×10^6 =x use crisscross

X= 260,000,000 ×240

X=6.24×10^10

3 0

3 years ago

Read 2 more answers

(1 point) The matrix A=⎡⎣⎢−4−4−40−8−4084⎤⎦⎥A=[−400−4−88−4−44] has two real eigenvalues, one of multiplicity 11 and one of multip

serious [3.7K]

Answer:

We have the matrix $A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right]$

To find the eigenvalues of A we need find the zeros of the polynomial characteristic $p(\lambda)=det(A-\lambda I_3)$

Then

$p(\lambda)=det(\left[\begin{array}{ccc}-4-\lambda&-4&-4\\0&-8-\lambda&-4\\0&8&4-\lambda\end{array}\right] )\\=(-4-\lambda)det(\left[\begin{array}{cc}-8-\lambda&-4\\8&4-\lambda\end{array}\right] )\\=(-4-\lambda)((-8-\lambda)(4-\lambda)+32)\\=-\lambda^3-8\lambda^2-16\lambda$

Now, we fin the zeros of $p(\lambda)$ .

$p(\lambda)=-\lambda^3-8\lambda^2-16\lambda=0\\\lambda(-\lambda^2-8\lambda-16)=0\\\lambda_{1}=0\; o \; \lambda_{2,3}=\frac{8\pm\sqrt{8^2-4(-1)(-16)}}{-2}=\frac{8}{-2}=-4$

Then, the eigenvalues of A are $\lambda_{1}=0$ of multiplicity 1 and $\lambda{2}=-4$ of multiplicity 2.

Let's find the eigenspaces of A. For $\lambda_{1}=0$ : $E_0 = Null(A- 0I_3)=Null(A)$ .Then, we use row operations to find the echelon form of the matrix

$A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right]\rightarrow\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&0&0\end{array}\right]$

We use backward substitution and we obtain

1.

$-8y-4z=0\\y=\frac{-1}{2}z$

2.

$-4x-4y-4z=0\\-4x-4(\frac{-1}{2}z)-4z=0\\x=\frac{-1}{2}z$

Therefore,

$E_0=\{(x,y,z): (x,y,z)=(-\frac{1}{2}t,-\frac{1}{2}t,t)\}=gen((-\frac{1}{2},-\frac{1}{2},1))$

For $\lambda_{2}=-4$ : $E_{-4} = Null(A- (-4)I_3)=Null(A+4I_3)$ .Then, we use row operations to find the echelon form of the matrix

$A+4I_3=\left[\begin{array}{ccc}0&-4&-4\\0&-4&-4\\0&8&8\end{array}\right] \rightarrow\left[\begin{array}{ccc}0&-4&-4\\0&0&0\\0&0&0\end{array}\right]$

We use backward substitution and we obtain

1.

$-4y-4z=0\\y=-z$

Then,

$E_{-4}=\{(x,y,z): (x,y,z)=(x,z,z)\}=gen((1,0,0),(0,1,1))$

8 0

3 years ago

PLEASE HELP WITH THIS MATH EQUATION!!!!

MA_775_DIABLO [31]

Answer:

53 bulbs

Step-by-step explanation:

3 0

4 years ago

Karmen returned a bicycle to Earl's Bike Shop. The sales receipt showed a total paid price of $211.86, including the 7% sales ta

MakcuM [25]

Answer:

$198

Step-by-step explanation:

198x.07=13.86

198+13.86=211.86

7 0

4 years ago

what is one possible value of x for which the function g above is undefined? g(x) = √(x-1) (x-2) note: √ iss the root sign

oksano4ka [1.4K]

Answer:

A possible value of x for which g(x) = √(x-1)(x-2) is undefined is x = 1.5

Step-by-step explanation:

The function g(x) = √(x-1)(x-2) is undefined when (x-1)(x-2) < 0.

So (x-1)(x-2) < 0

⇒ x - 1 < 0 or x -2 < 2

⇒ x < 1 or x < 2.

So we look for the interval for which (x - 1)(x - 2) < 0

when x < 1, e.g x = 0, (x - 1)(x - 2) = (0 - 1)(0 - 2) = (-1)(-2) =2 > 0. So (x - 1)(x - 2) > 0 for x < 1

when 1 < x < 2, e.g x = 1.5, (x - 1)(x - 2) = (1.5 - 1)(1.5 - 2) = (0.5)(-0.5) = -0.25 < 0. So (x - 1)(x - 2) < 0 for 1 < x < 2

when x > 2 e.g x = 3, (x - 1)(x - 2) = (3 - 1)(3 - 2) = (2)(1) = 2 > 0. So (x - 1)(x - 2) > 0 for x > 2

Since (x - 1)(x - 2) < 0 for 1 < x < 2, this is the required interval.

So, g(x) = √(x-1)(x-2) is undefined in the interval 1 < x < 2

A possible value for x in this interval is x = 1.5

So, a possible value of x for which g(x) = √(x-1)(x-2) is undefined is x = 1.5

5 0

4 years ago