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3 years ago

M is the midpoint of CF (line over it) for the points C(3,4) and F(9,8). Find MF

1 answer:

3 0

$(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) \\(\frac{3 + 9}{2}, \frac{4 + 8}{2}) \\(\frac{12}{2}, \frac{12}{2}) \\(6, 6)$

Midpoint of CF: (6, 6)

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Solve the system below using elimination {5x+9y= -10} {7x+10y= -1}

bixtya [17]

∑Hey , jillianwagler ⊃

Answer:

x = 7, y = -5

Step-by-step explanation:

Given~:

Solve the system below using elimination

{5x+9y= -10}

{7x+10y= -1}

Solve:

$\begin{bmatrix}5x+9y=-10\\ 7x+10y=-1\end{bmatrix}$

Isolate x ( 5x + 9y = -10 ) : $\mathrm{x=\frac{-10-9y}{5}}$

Substitute: $\mathrm{x=\frac{-10-9y}{5}}$

$\begin{bmatrix}7\cdot \frac{-10-9y}{5}+10y=-1\end{bmatrix}$

$\begin{bmatrix}\frac{-70-13y}{5}=-1\end{bmatrix}$

Isolate y : $\frac{-70-13y}{5}=1:y=-5$

Substitute: $y=-5$

$x=\frac{-10-9\left(-5\right)}{5}=7$

Hence, $x=7$

Therefore, the solution for $\begin{bmatrix}5x+9y=-10\\ 7x+10y=-1\end{bmatrix}$ are:

$x=7,\:y=-5$

xcookiex12

8/19/2022

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2 years ago

Would you use the median or mean to describe the center of each data set? Explain your reasoning. Heights of 50 basketball playe

Ludmilka [50]

Answer:

Median is used to describe the center of the data of heights of 50 basketball players.

Step-by-step explanation:

The median divides the data into two equal parts.

It can be used to describe the center of the data.

The mean gives the average where most data lies.

The advantages of median are

1)it is located even when the values are not capable of quantitative measurements.

2) It is not affected by extreme values > It can be computed even when a frequency distribution involves open end classes like those of income and prices.

3) In a highly skewed distribution median is appropriate average to use.

The disadvantages of the mean are

1) it is greatly affected by extreme values.

2) It sometimes gives fallacious conclusions.

3) In a highly skewed distribution mean is not an appropriate average to use.

4) It cannot be computed even when a frequency distribution involves open end classes like those of income and prices.

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3 years ago

Tim's bean sprout grew 3/ inches. Teegan's bean sprout grew 2 3/4 inches. How many more inches did Tim's bean sprout grow than T

ki77a [65]

Answer:

Tim's bean sprout grow by more than $\frac{5}{8} \text{ inches}$ .