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swat32
4 years ago
8

Seven times a number minus the number is -48. Find the number.

Mathematics
2 answers:
yKpoI14uk [10]4 years ago
6 0
The answer youre looking for is -8!
Wittaler [7]4 years ago
3 0
Remark
Set up the equation from the word problem.

Equation
7*x - x = -48  Combine like terms 
6x = - 48 Divide by 6
x = - 48 / 6
x = - 8 <<<<<  Answer
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)) Omar used 28 centimeters of tape to wrap 4 presents. How much tape will Omar need in
pav-90 [236]

Answer:

He would need 42 cm *or 14cm

Step-by-step explanation:

28/4 is 7

that means that he used 7 cm of tape per present

6 x 7 is 42

that means that he would need 42 cm of tape

4 to 28

6 to 42

*If he already has 4 presents done and he needs to 2 more present then he would need 14 cm, 2 x 7 is 14

8 0
4 years ago
Match the statement to its correct expression.
Mashutka [201]
Oh my god you have got to be joking.
C IS OBVIOUSLY THE ANSWER
Eight more is + 8, and the quotient of a number and four would be n <span>÷ 4
</span>
6 0
4 years ago
(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
I need some help please
klio [65]
True, because the dilation factor is less than one. This will result in the shape reducing.
4 0
4 years ago
Which graph represents the solution to the system of equations?
Sauron [17]

Answer: bottom right

Step-by-step explanation:

5 0
3 years ago
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