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timama [110]
3 years ago
8

A triangle is isosceles is the altitude from the vertex angle to the base also passes through the midpoint of the base. What are

the two slopes from the figure that prove this? M is the midpoint of triangle ABC.
-7 and 1/7

5 and -1/5

-5 and 1/5

7 and -1/7

Mathematics
1 answer:
leonid [27]3 years ago
5 0

Answer:

The answer to your question is  7 and -1/7

Step-by-step explanation:

Data

First line                        Second line

B (4, 4)                            A (0, 1)

M (3.5, 0.5)                     C (7, 0)

Formula

Slope = m = (y2 - y1) / (x2 - x1)

Process

1.- Calculate the slope of the first-line

    x1 = 4        y1 = 4

    x2 = 3.5    y2 = 0.5

-Substitution

    m = (0.5 - 4) / (3.5 - 4)

-Simplification

     m = -3.5 / -0.5

-Result

     m = 7

2.- Calculate the slope of the second line

    x1 = 0          y1 = 1

    x2 = 7         y2 = 0

-Substitution

   m = (0 - 1) / (7 - 0)

-Simplification

   m = -1/7

-Result

   m = -1/7  

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This week, we are covering relationships that can be approximated by linear equations. For instance, y = 453x + 3768 represents
lana [24]

Answer:

See explanation below.

Step-by-step explanation:

We assume that the data is given by :

x: 30, 30, 30, 50, 50, 50, 70,70, 70,90,90,90

y: 38, 43, 29, 32, 26, 33, 19, 27, 23, 14, 19, 21.

Where X represent the cost for scholarships in thousands of dollars and y represent the cost of life for an academic semester (The data comes from the web)

We can find the least-squares line appropriate for this data.  

For this case we need to calculate the slope with the following formula:

m=\frac{S_{xy}}{S_{xx}}

Where:

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}

So we can find the sums like this:

\sum_{i=1}^n x_i = 30+30+30+50+50+50+70+70+70+90+90+90=720

\sum_{i=1}^n y_i =38+43+29+32+26+33+19+27+23+14+19+21=324

\sum_{i=1}^n x^2_i =30^2+30^2+30^2+50^2+50^2+50^2+70^2+70^2+70^2+90^2+90^2+90^2=49200

\sum_{i=1}^n y^2_i =38^2+43^2+29^2+32^2+26^2+33^2+19^2+27^2+23^2+14^2+19^2+21^2=9540

\sum_{i=1}^n x_i y_i =30*38+30*43+30*29+50*32+50*26+50*33+70*19+70*27+70*23+90*14+90*19+90*21=17540

With these we can find the sums:

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=49200-\frac{720^2}{12}=6000

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}=17540-\frac{720*324}{12}{12}=-1900

And the slope would be:

m=-\frac{1900}{6000}=-0.317

Nowe we can find the means for x and y like this:

\bar x= \frac{\sum x_i}{n}=\frac{720}{12}=60

\bar y= \frac{\sum y_i}{n}=\frac{324}{12}=27

And we can find the intercept using this:

b=\bar y -m \bar x=27-(-0.317*60)=46.02

So the line would be given by:

y=-0.317 x +46.02

We have an inverse linear relationship since the slope is negative between the variables of interest.

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