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zimovet [89]
3 years ago
8

There are 90 girls and 60 boys in the sixth grade at a middle school. Of these students, 9 girls and 3 boys write left-handed. W

hat percentage of the sixth graders at this middle school write left handed?
Mathematics
1 answer:
Juliette [100K]3 years ago
6 0
.08% 90+60=150 and 9+3=12 12/150=.08
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4a + 6b= 10 2a - 4b =12 what does 12a=
mixer [17]
4a + 6b = 10
2a - 4b = 12...multiply by -2
----------------
4a + 6b = 10
-4a + 8b = - 24 (result of multiplying by -2)
------------------add
14b = - 14
b = -14/14
b = -1

2a - 4b = 12
2a - 4(-1) = 12
2a + 4 = 12
2a = 12 - 4
2a = 8
a = 8/2
a = 4

so 12a = 12(4) = 48 <==
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What is 18/32 in simplest form​
JulsSmile [24]

Answer:

Step-by-step explanation:

18/32 is the same as 9/16 or 0.5625

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Please help with this question thanksm
cluponka [151]

Answer: a) 1/64

Step-by-step explanation:

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3 years ago
Circle D is below what is the arc measure of BC in degrees
maks197457 [2]

Answer:

84°

Step-by-step explanation:

\because (13n - 16) \degree + (7n  + 12) \degree + (6n)\degree \\= 360\degree \\  \\  \therefore \: (26n - 4) \degree= 360\degree \\  \\  \therefore \: 26n - 4= 360 \\  \\ \therefore \: 26n = 360 + 4 \\  \\ \therefore \: 26n = 364 \\  \\ \therefore \: n =  \frac{364}{26}  \\  \\ \therefore \: n =  14 \\  \\\because  m\overset { \frown}{BC} = (6n)°\\  \\\therefore \:m \overset { \frown}{BC} =(6\times 14)°  \\  \\   \huge \purple {\boxed {\therefore \:m\overset { \frown}{BC}= 84°}} \\

4 0
3 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] =<em>I</em>.

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

<em>I</em> = \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
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