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kotykmax [81]
4 years ago
5

Simplify 12x to the power of 7 y to the power of 3 divided by 6x to the power of 3 y

Mathematics
1 answer:
Galina-37 [17]4 years ago
7 0

Answer:

2x⁴y²

Step-by-step explanation:

The question gives us the expression;

12x⁷y³ ÷ 6x³y

We are supposed to simplify the expression;

First we could divide the digits;

12 ÷ 6 = 2

Then we divide the like terms;

x⁷ ÷ x³

This means;

=\frac{x*x*x*x*x*x*x }{x*x*x}

= x*x*x*x

=x^4

y³ ÷ y

That is;

=\frac{y*y*y}{y}

=y*y

=y^2

Therefore, the simplified expression will be 2x⁴y²

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4 0
3 years ago
Rewrite the system of linear equations as a matrix equation AX = B.
iren2701 [21]

Answer:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

Equations:

Eq. 1 : x1 + 2x2 + 5x3 = 5

Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

Coefficients for x1 ; x2 ; x3

From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]

And the vector of vriables (X) is:

\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

In conclusion, AX=B is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

7 0
4 years ago
I need help w/ this!! thank you
Fittoniya [83]
<h3><u>Answer:</u></h3>

\boxed{\pink{\sf \leadsto Yes \ there \ is \ a \ solution \ of \ the \ given \ inequality .}}

<h3><u>Step-by-step explanation:</u></h3>

A inequality is given to us and we need to convert it into standard form and see whether if it has a solution . So let's solve the inequality.

The inequality given to us is :-

\bf\implies |2y + 3 | - 1 \leq 0 \\\\\bf\implies |2y+3|\leq 1 \\\\\bf\implies (|2y+3|)^2 \leq 1^2  \\\\\bf\implies (2y+3)^2 \leq 1  \\\\\bf\implies (2y)^2+3^2+2(2y)(3) \leq 1  \\\\\bf\implies 4y^2+9+12y - 1 \leq 0  \\\\\bf\implies 4y^2+12y+8 \leq 0 \\\\\bf\implies 4( y^2 + 3y + 2 ) \leq 0  \\\\\bf\implies y^2+3y +2 \leq 0 \:\:\bigg\lgroup \purple{\bf Standard \ form \ of \ inequality }\bigg\rgroup   \\\\\bf\implies y^2y+2y+y+2 \leq 0  \\\\\bf\implies y(y+2)+1(y+2)\leq 0  \\\\\bf\implies ( y+2)(y+1)\leq 0  \\\\\bf\implies \boxed{\red{\bf y \leq (-2) , (-1) }}

Let's plot a graph to see its interval . Graph attached in attachment .

Now we can see that the Interval notation of would be ,

\boxed{\boxed{\orange \tt \purple{\leadsto}y \in [-2,-1] }}

<h3><u>Hence</u><u> the</u><u> </u><u>standa</u><u>rd</u><u> </u><u>form</u><u> </u><u>of</u><u> </u><u>inequa</u><u>lity</u><u> </u><u>is</u><u> </u><u>y²</u><u>+</u><u>3y</u><u> </u><u>+</u><u>2</u><u> </u><u>≤</u><u> </u><u>0</u><u> </u><u>and</u><u> </u><u>the </u><u>Solution</u><u> </u><u>set</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>ineq</u><u>uality</u><u> </u><u>is</u><u> </u><u>[</u><u> </u><u>-</u><u>2</u><u> </u><u>,</u><u> </u><u>-</u><u>1</u><u> </u><u>]</u><u> </u><u>.</u></h3>
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2 years ago
N − 7.32 / 2 = 1.04<br><br> What is "n"?
Pachacha [2.7K]

Answer:

n= 4.7

Step-by-step explanation:

n+−3.66=1.04

n−3.66=1.04

Step 2: Add 3.66 to both sides.

n−3.66+3.66=1.04+3.66

n=4.7

Answer:

n=4.7

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3 years ago
Romeo will roll a 6-sided number cube, numbered 1-6, twice. What is the probability of rolling an odd number on the 1st roll and
elena-s [515]
1st rol= 3/6
2nd roll= 1/6
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