All categories

3 years ago

Sally has a string that has a length of 13 inches. She used the string to make two bows.

Mathematics

1 answer:

olganol [36]3 years ago

7 0

The string is 6 inch

You might be interested in

The net of a prism is shown on the coordinate plane. What is the surface area of the prism?

nata0808 [166]

Answer:

Option C. 10 square inch

Step-by-step explanation:

Net of a prism is shown on the coordinate plane. We have to calculate the surface area of the prism.

To calculate the surface area of the net we will calculate the area of the four large rectangles and two squares given in the picture.

Total surface area of the net = 2× small squares + 4×large rectangles

= 2×(1×1) + 4×(1×2) = 2 + 8 = 10 square inch.

Therefore option C. 10 square inch is the answer.

6 0

3 years ago

41/25 as a percentage in math ???

Anarel [89]

165% , is your answer.
41 divided by 25 is 1.65 times 100.

8 0

3 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

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

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

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

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

Then,

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

8 0

3 years ago

How we solve a b and c

Studentka2010 [4]

For a. 55greater than equal to X less than equal to 80

8 0

3 years ago

Which expression is equivalent to 4 sqrt 16x^11y^8/81x^7y^6

DanielleElmas [232]

The first step for finding out whether or not this expression is equivalent to $\sqrt{4}$ is to reduce the fraction with $x^{7}$ . You can begin to do this by dividing the terms with the same base by subtracting their exponents.
$\frac{16 x^{11-7} y^{8} }{81 y^{6} }$
Subtract the exponents.
$\frac{16 x^{4} y^{8} }{81 y^{6} }$
Now reduce the fraction with $y^{6}$ by doing the same process. Since I just showed you how to do this,, I will skip over this.
$\frac{16 x^{4} y^{2} }{81}$
Since we cannot simplify this expression any further,, your answer is going to be $\frac{16 x^{4} y^{2} }{81}$ ,, which is not equivalent to $\sqrt{4}$ .
Let me know if you have any further questions.
:)

8 0

3 years ago