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3 years ago

A student is preparing an electronic slide presentation about high school athletics. Which multimedia resources

Mathematics

2 answers:

Katarina [22]3 years ago

7 0

Answer:

Step-by-step explanation:

5 & 6

evablogger [386]3 years ago

3 0

Answer:

5 and 6

Step-by-step explanation:

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Pls help question in image

cluponka [151]

Answer: 41.16 cm cubed.

= (0.5 • 4.2 • 2.8) • 7 = 41. 16

Explanation: B = 1/2 base times hight of the triangle. Not the figure.
Remember the figure is also tipped over, and its not up right. So you’ll have to try to imagine and image it standing up right.

6 0

1 year ago

Using distributive property what is 3(5 +6) =

Annette [7]

Answer:

3(5 +6) = 33

Step-by-step explanation:

3(5+6)

so first you multiply 3 by 5

which is 15

then you multiply 3 by 6

which is 18

and that remains as 15+18

which in total equals 33

7 0

2 years ago

Read 2 more answers

of this data set 56, 57, 57, 58, 59, 60, 60, 60, 62, 64, 64, 66, 67, 67, 67, 68 what's the mean absolute deviation

seropon [69]

Answer:

3.625

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

HELP PLZ Given that ABEF⩭CBEF,m∠BED=82°,and m∠CBE=152°,find m∠A (Hint: The angles in a quadrilateral=360°)

lesya692 [45]

(A) 28 . Thanks for the question hope this helps

6 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

3 years ago