All categories

3 years ago

rosa correctly solved 275 divided by 5.enter numbers in boxes to show the equations she could use to solve the problem

Mathematics

1 answer:

kifflom [539]3 years ago

6 0

Answer:

The first step would be to look at the first two numbers (which is 27) and estimate which multiple of 5 is closest and below 27.

5*5 = 25

So once she got 25, she will need to subtract 27-25, which will give her 2. REMEMBER: 5 is the first number of the quotient.

Now, she will need to drag the last digit left of 275 (which is 5) to the remainder 2 and think what multiple of 5 will give her the answer 25.

Again, 5*5 = 25

Once again, her numbers on the quotient will be 55. That's the answer.

You might be interested in

13=b 5=c what is the length of the missing leg

svetoff [14.1K]

Answer:

The length of the missing leg is 12

Step-by-step explanation:

Right Triangles

A right triangle is a triangle that has an internal angle of 90°. In right triangles, the Pythagorean relation between its side length stands.

Suppose a and c are the smaller sides, also called 'legs' of the triangle, and b is the bigger side, also known as hypotenuse, then:

$a^2+c^2=b^2$

We are given the hypotenuse b=13 and one leg c=5. Let's find the missing leg solving the above equation for a:

$a^2=b^2-c^2$

Substituting:

$a^2=13^2-5^2$

$a^2=169-25=144$

Solving:

$a=\sqrt{144}=12$

The length of the missing leg is 12

7 0

4 years ago

A truck costs $80,000. It depreciates in value $6,000 per year. Write a linear model in the form v(t)=mt + b where v(t) represen

zheka24 [161]

Answer:

v(t)=(-6,000)(years)+80,000

Step-by-step explanation:

4 0

4 years ago

32.6 × 10<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20x%5E%7B2%7D%20%5C%5Cx%5E%7B2%7D" id="TexFormula1" title="x^{2} x^{2} \\

borishaifa [10]

Answer:

A plants grow form a ?

Seed

4 0

3 years ago

Consider the matrix A. A = 1 0 1 1 0 0 0 0 0 Find the characteristic polynomial for the matrix A. (Write your answer in terms of

dusya [7]

Answer with Step-by-step explanation:

We are given that a matrix

$A=\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]$

a.We have to find characteristic polynomial in terms of A

We know that characteristic equation of given matrix $\mid{A-\lambda I}\mid=0$

Where I is identity matrix of the order of given matrix

I= $\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$

Substitute the values then, we get

$\begin{vmatrix}1-\lambda&0&1\\1&-\lambda&0\\0&0&-\lambda\end{vmatrix}=0$

$(1-\lambda)(\lamda^2)-0+0=0$

$\lambda^2-\lambda^3=0$

$\lambda^3-\lambda^2=0$

Hence, characteristic polynomial = $\lambda^3-\lambda^2=0$

b.We have to find the eigen value for given matrix

$\lambda^2(1-\lambda)=0$

Then , we get $\lambda=0,0,1-\lambda=0$

$\lambda=1$

Hence, real eigen values of for the matrix are 0,0 and 1.

c.Eigen space corresponding to eigen value 1 is the null space of matrix $A-I$

$E_1=N(A-I)$

$A-I=\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&-1\end{array}\right]$

Apply $R_1\rightarrow R_1+R_3$

$A-I=\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&0\end{array}\right]$

Now,(A-I)x=0[/tex]

Substitute the values then we get

$\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&0\end{array}\right]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=0$

Then , we get $x_3=0$

And $x_1-x_2=0$

$x_1=x_2$

Null space N(A-I) consist of vectors

x= $\left[\begin{array}{ccc}x_1\\x_1\\0\end{array}\right]$

For any scalar $x_1$

$x=x_1\left[\begin{array}{ccc}1\\1\\0\end{array}\right]$

$E_1=N(A-I)=Span(\left[\begin{array}{ccc}1\\1\\0\end{array}\right]$

Hence, the basis of eigen vector corresponding to eigen value 1 is given by

$\left[\begin{array}{ccc}1\\1\\0\end{array}\right]$

Eigen space corresponding to 0 eigen value

$N(A-0I)=\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]$

$(A-0I)x=0$

$\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=0$

$\left[\begin{array}{ccc}x_1+x_3\\x_1\\0\end{array}\right]=0$

Then, $x_1+x_3=0$

$x_1=0$

Substitute $x_1=0$

Then, we get $x_3=0$

Therefore, the null space consist of vectors

$x=x_2=x_2\left[\begin{array}{ccc}0\\1\\0\end{array}\right]$

Therefore, the basis of eigen space corresponding to eigen value 0 is given by

$\left[\begin{array}{ccc}0\\1\\0\end{array}\right]$

5 0

3 years ago

The circumference of a circle is 157 centimeters. What is the area of the circle in terms

Kruka [31]

The answer is B. I need 20 characters so sorry, jwdsausvndjsdvjvsddcjbcdjadcnahadcjwdn

6 0

3 years ago