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olasank [31]
3 years ago
14

6x + 3 - 6x = 3

Mathematics
2 answers:
nata0808 [166]3 years ago
8 0
Answer:
B. No solution


Step by step explanation:
6x+3-6x=3
6x-6x=0
0=0
Elena L [17]3 years ago
5 0

Answer:

0

Step-by-step explanation:

all real numbers are solutions

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What is the VALUE OF X, for question one? Show work if possible
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In a parallelogram, the diagonals bisect each other.

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3 years ago
Answer asap please
Oduvanchick [21]

Answer:

(0,1) and (-2,9)

Step-by-step explanation:

A point (x,y) lies on the graph of f(x) = (\frac{1}{3} )^{x} if it satisfies the condition y = (\frac{1}{3} )^{x}

<h3>(0,1):</h3>

        x = 0 and y = 1.

This point satisfies the required condition as

        (\frac{1}{3} )^{0} = 1

Hence, this point <u>is</u> on the graph of f(x) = (\frac{1}{3} )^{x}

<h3>(3,27):</h3>

        x = 3 and y = 27.

This point doesn't satisfy the required condition as

        (\frac{1}{3} )^{3} = \frac{1}{27} ≠ 27

Hence, this point i<u>s not</u> on the graph of f(x) = (\frac{1}{3} )^{x}

<h3>(-2,9):</h3>

        x = -2 and y = 9.

This point satisfies the required condition as

        (\frac{1}{3} )^{-2} = 3^{2} = 9

Hence, this point <u>is</u> on the graph of f(x) = (\frac{1}{3} )^{x}

<h3>(-1,-\frac{1}{3}):</h3>

        x = -1 and y = -\frac{1}{3}.

This point satisfies the required condition as

        y = (\frac{1}{3} )^{-1} = 3^{1} = 3 ≠ -\frac{1}{3}

Hence, this point <u>is not</u> on the graph of f(x) = (\frac{1}{3} )^{x}

7 0
3 years ago
What is the value of -16?
timama [110]

Answer:

the answer would be -4

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
A = ???? 4 −2
irinina [24]

Answer:

1. The matrix A isn't the inverse of matrix B.

2. |B|=12, |A|=12

Step-by-step explanation:

1. We want to know if matrix A is the inverse of matrix B, this means that if you do the product between B and A you have to obtain the identity matrix.

We have:

A=\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right]

and

B=\left[\begin{array}{cc}3&2\\1&4\end{array}\right]

A and B are 2×2 matrices (2 rows and 2 columns), if you multiply them you have to obtain a 2×2 matrix.

Then if A is the inverse of B:

B.A=I

Where,

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

Observation:

If you have two matrices:

A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\and\\B=\left[\begin{array}{cc}e&f\\g&h\end{array}\right]\\\\\\A.B=\left[\begin{array}{cc}(a.e+b.g)&(a.f+b.h)\\(c.e+d.g)&(c.f+d.h)\end{array}\right]

Now:

B.A=\left[\begin{array}{cc}3&2\\1&4\end{array}\right].\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}4.3+(-2).1&4.2+(-2).4\\(-1).3+3.1&(-1).2+3.4\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}12-2&8-8\\-3+3&-2+12\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}10&0\\0&10\end{array}\right]

B.A=\left[\begin{array}{cc}10&0\\0&10\end{array}\right]\neq \left[\begin{array}{cc}1&0\\0&1\end{array}\right]=I\\\\\\B.A\neq I

Then, the matrix A isn't the inverse of matrix B.

2. If you have a matrix A:

A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]

The determinant of the matrix is:

|A|=ad-bc

Then the determinant of B is:

B=\left[\begin{array}{cc}3&2\\1&4\end{array}\right]

a=3, b=2, c=1, d=4

|B|=3.4-2.1\\|B|=12-2=10

The determinant of A is:

A=\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right]

a=4, b=-2, c=-1, d=3

|A|=4.3-(-2).(-1)\\|B|=12-2=10

6 0
4 years ago
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