All categories

2 years ago

would you choose to use a hundred chart to solve 90- 80? why or why. not ? if not. which strategy would work better?

1 answer:

5 0

I would use a tens chart because both numbers are multiples of 10, and there both under 100, so it would work out better if u use a tens chart, since the numbers are in the tens place and would give u a product of 10.

i hope this helps u

You might be interested in

Mr. Mutler posts his students' artwork on a bulletin board.

Jlenok [28]

The length could be 5 and width 3. That would equal 15

8 0

2 years ago

Solve for XX. Assume XX is a 2×22×2 matrix and II denotes the 2×22×2 identity matrix. Do not use decimal numbers in your answer.

sveticcg [70]

The question is incomplete. The complete question is as follows:

Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.

$\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ =I.

First, we have to identify the matrix I. As it was said, the matrix is the identiy matrix, which means

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

So, $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Isolating the X, we have

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ - $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Resolving:

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]$

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,

X= $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ ⁻¹· $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.

So,

$\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ · $\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

9a - 3b = 1

7a - 6b = 0

9c - 3d = 0

7c - 6d = 1

Resolving these equations, we have a= $\frac{2}{11}$ ; b= $\frac{7}{33}$ ; c= $\frac{-1}{11}$ and d= $\frac{-3}{11}$ . Substituting:

X= $\left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11} \end{array}\right]$ · $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Multiplying the matrices, we have

X= $\left[\begin{array}{ccc}\frac{8}{11} &\frac{26}{11} \\\frac{39}{11}&\frac{198}{11} \end{array}\right]$

6 0

2 years ago

Please helm me

spin [16.1K]

Area of circular metal=πr^2 =3.14 x 3.5 x 3.5 m^2 now, total cost=area x rate =3.14 x 3.5 x 3.5 x 3.25 $ =125.01 $

4 0

2 years ago

Read 2 more answers

Which line segment is drawn in the figure? A.YZ B.WY C.XZ D.WZ

Dima020 [189]

The line segment shown in the figure is C. XZ. A line segment is a line that is bounded by 2 endpoints, hence the line is not infinite. As shown in the figure, only points X and Z are visibly connected. There are no visible lines connecting the other points together.

6 0

2 years ago

Read 2 more answers

In Friday NIght Funkin' dave and bambi How and What did Dave become 3D?

Ierofanga [76]

Answer:

Step-by-step explanation:

because you rupedid lol

5 0

1 year ago