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Mashcka [7]
3 years ago
12

The point (3,2) reflected in the x-axis is (

Mathematics
1 answer:
ioda3 years ago
8 0

Answer:

reflected across the x-axis is 3,-2 and the y axis is -3,2

Step-by-step explanation:

When you reflect over the X axis it will stay the same numbers but change the Y axis sign. the same thing happens to the y -axis but the x changes siogns instead

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The table shows a propotional realationship between x and y
ss7ja [257]

Answer:

jvgdvfysvj fjrfhlgd ghcfh ghgdc

vurv gshgb gjjdvk jfdsuk0v

6 0
2 years ago
What is 5/6 and 3/8 as an equivalent fraction ​
Veronika [31]

Answer:

5/6 - 10/12, 15/18, 20/24

3/8 - 6/16, 9/24, 12/32

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
What happens to the mode of the data set shown below if
nlexa [21]

Answer:

OB) The mode does not change

8 0
3 years ago
For waht values of x do the vectors -1,0,-1), (2,1,2), (1,1, x) form a basis for R3?
DaniilM [7]
<h2>Answer:</h2>

The values of x for which the given vectors are basis for R³ is:

                        x\neq 1

<h2>Step-by-step explanation:</h2>

We know that for a set of vectors are linearly independent if the matrix formed by these set of vectors is non-singular i.e. the determinant of the matrix formed by these vectors is non-zero.

We are given three vectors as:

(-1,0,-1), (2,1,2), (1,1, x)

The matrix formed by these vectors is:

\left[\begin{array}{ccc}-1&2&1\\0&1&1\\-1&2&x\end{array}\right]

Now, the determinant of this matrix is:

\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-1(x-2)-2(1)+1\\\\\\\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-x+2-2+1\\\\\\\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-x+1

Hence,

-x+1\neq 0\\\\\\i.e.\\\\\\x\neq 1

4 0
3 years ago
Water whose temperature is at 100∘C is left to cool in a room where the temperature is 60∘C. After 3 minutes, the water temperat
Tju [1.3M]

Answer:

21.68 minutes ≈ 21.7 minutes

Step-by-step explanation:

Given:

T=60+40e^{kt}

Initial temperature

T = 100°C

Final temperature = 60°C

Temperature after (t = 3 minutes) = 90°C

Now,

using the given equation

T=60+40e^{kt}

at T = 90°C and  t = 3 minutes

90=60+40e^{k(3)}

30=40e^{3k}

or

e^{3k}=\frac{3}{4}

taking the natural log both sides, we get

3k = \ln(\frac{3}{4})

or

3k = -0.2876

or

k = -0.09589

Therefore,

substituting k in 1 for time at temperature, T = 65°C

65=60+40e^{( -0.09589)t}

or

5=40e^{( -0.09589)t}

or

e^{( -0.09589)t}=\frac{5}{40}

or

e^{( -0.09589)t}=0.125

taking the natural log both the sides, we get

( -0.09589)t = ln(0.125)

or

( -0.09589)t = -2.0794

or

t = 21.68 minutes ≈ 21.7 minutes

6 0
3 years ago
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