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vichka [17]
4 years ago
7

3x+1=7x+8 Solve the equation

Mathematics
1 answer:
pychu [463]4 years ago
5 0

Answer:

x=-7/4

3x+1=7x+8

-3x     -3x

1=      4x+ 8

-8            -8

-7/4=x

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The answer to the question is D
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Three vertices of a parallelogram are shown in the figure below.
a_sh-v [17]

Given:

The three vertices of the parallelogram are (-3,9), (0,-3), (6,-6).

To find:

The fourth vertex of the parallelogram.

Solution:

Consider the three vertices of the parallelogram are A(-3,9), B(0,-3), C(6,-6).

Let D(a,b) be the fourth vertex.

Midpoint formula:

Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)

We know that the diagonals of a parallelogram bisect each other. So, the midpoints of both diagonals are same.

Midpoint of AC = Midpoint BD

\left(\dfrac{-3+6}{2},\dfrac{9+(-6)}{2}\right)=\left(\dfrac{0+a}{2},\dfrac{-3+b}{2}\right)

\left(\dfrac{3}{2},\dfrac{3}{2}\right)=\left(\dfrac{a}{2},\dfrac{-3+b}{2}\right)

On comparing both sides, we get

\dfrac{3}{2}=\dfrac{a}{2}

3=a

And,

\dfrac{3}{2}=\dfrac{-3+b}{2}

3=-3+b

3+3=b

6=b

Therefore, the fourth vertex of the parallelogram is (3,6).

7 0
3 years ago
Help help help help help PLEASE
Alinara [238K]
A. 0,0 1,14. It is showing that it’s a positive correlation
B. I think 120
4 0
3 years ago
Read 2 more answers
Use the box plots comparing the number of males and number of females attending the latest superhero movie each day for a month
Maksim231197 [3]

Answer:

Part A:

The interquartile range is approximately 10

Part B:

The difference between the median values for each data set  is approximately 6

Part C:

i) More widely distributed and concentrated to the beginning of the month

The better measure of the center for the male dataset is the median

ii) The skewed distribution

The mean is the better measure of center for the dataset

Part D;

A possible reason for the outlier is by chance

Step-by-step explanation:

Part A:

The interquartile range, IQR, is the width of the box in the box plot = The difference between the third quartile, Q₃ and the first quartile, Q₁ = Q₃ - Q₁

From the diagram, Q₃ ≈ 13, Q₁ ≈ 3

∴ The interquartile range, IQR = 13 - 3 = 10

Part B:

From the diagram of the box plot, the median value, of the males, M-Q₂ ≈ 12

The median value for the females, F-Q₂ ≈ 18

The difference, M-Q₂ - F-Q₂ = 18 - 12 ≈ 6

The difference between the median values for each data set, d = M-Q₂ - F-Q₂ ≈ 6

Part C:

i) The distribution of the of the male dataset for the male has more data on the left of the box plot, with the median located approximately at the center and an outlier at end of the month

The better measure of the center for the male dataset is the median

ii) The distribution of the dataset for the females is skewed (concentrated on the right of the plot). The data set has no outlier and the first quartile is 0, while the third quartile corresponds with the maximum value at 21, the first quartile is 15 and the IQR is approximately 6

The better measure of center for the dataset is the mean, as the difference between the value of the first quartile, 0, and the other data points is more accounted for by the mean

Part D;

The possible reason for the outlier in the male dataset can be attributed to chance

4 0
3 years ago
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