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schepotkina [342]

3 years ago

15

The chart shows Daniela’s run through her race.

1 answer:

Nutka1998 [239]3 years ago

8 0

So it would be c the

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What is the length of side s of the square shown below?

yarga [219]

Answer:5 sqr 2

Step-by-step explanation:

4 0

3 years ago

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oksian1 [2.3K]

I think you are right since 2 plus 2 equals 4

3 0

3 years ago

Plz help ASAP!!!!!!!!!

N76 [4]

2. angle one congruent to angle two
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6 0

3 years ago

If A(4 -6) B(3 -2) and C (5 2) are the vertices of a triangle ABC fine the length of the median AD from A to BC. Also verify tha

Gnoma [55]

Answer:

a) The median AD from A to BC has a length of 6.

b) Areas of triangles ABD and ACD are the same.

Step-by-step explanation:

a) A median is a line that begin in a vertix and end at a midpoint of a side opposite to vertix. As first step the location of the point is determined:

$D (x,y) = \left(\frac{x_{B}+x_{C}}{2},\frac{y_{B}+y_{C}}{2} \right)$

$D(x,y) = \left(\frac{3 + 5}{2},\frac{-2 + 2}{2} \right)$

$D(x,y) = (4,0)$

The length of the median AD is calculated by the Pythagorean Theorem:

$AD = \sqrt{(x_{D}-x_{A})^{2}+ (y_{D}-y_{A})^{2}}$

$AD = \sqrt{(4-4)^{2}+[0-(-6)]^{2}}$

$AD = 6$

The median AD from A to BC has a length of 6.

b) In order to compare both areas, all lengths must be found with the help of Pythagorean Theorem:

$AB = \sqrt{(x_{B}-x_{A})^{2}+ (y_{B}-y_{A})^{2}}$

$AB = \sqrt{(3-4)^{2}+[-2-(-6)]^{2}}$

$AB \approx 4.123$

$AC = \sqrt{(x_{C}-x_{A})^{2}+ (y_{C}-y_{A})^{2}}$

$AC = \sqrt{(5-4)^{2}+[2-(-6)]^{2}}$

$AC \approx 4.123$

$BC = \sqrt{(x_{C}-x_{B})^{2}+ (y_{C}-y_{B})^{2}}$

$BC = \sqrt{(5-3)^{2}+[2-(-2)]^{2}}$

$BC \approx 4.472$

$BD = CD = \frac{1}{2}\cdot BC$ (by the definition of median)

$BD = CD = \frac{1}{2} \cdot (4.472)$

$BD = CD = 2.236$

$AD = 6$

The area of any triangle can be calculated in terms of their side length. Now, equations to determine the areas of triangles ABD and ACD are described below:

$A_{ABD} = \sqrt{s_{ABD}\cdot (s_{ABD}-AB)\cdot (s_{ABD}-BD)\cdot (s_{ABD}-AD)}$ , where $s_{ABD} = \frac{AB+BD+AD}{2}$

$A_{ACD} = \sqrt{s_{ACD}\cdot (s_{ACD}-AC)\cdot (s_{ACD}-CD)\cdot (s_{ACD}-AD)}$ , where $s_{ACD} = \frac{AC+CD+AD}{2}$

Finally,

$s_{ABD} = \frac{4.123+2.236+6}{2}$

$s_{ABD} = 6.180$

$A_{ABD} = \sqrt{(6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6)}$

$A_{ABD} \approx 3.004$

$s_{ACD} = \frac{4.123+2.236+6}{2}$

$s_{ACD} = 6.180$

$A_{ACD} = \sqrt{(6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6)}$

$A_{ACD} \approx 3.004$

Therefore, areas of triangles ABD and ACD are the same.

4 0

4 years ago

How to find the median

Leya [2.2K]

Answer:

How to Find the Median

Step-by-step explanation:

To find the Median of a number you can put the numbers in order and cross each one until you get a one number

Example:

1,2,3,4,<em>5</em>

The Number 3 that is bolded will be the Median

But if there is two numbers left which you can't cross out

you will have to divide those two numbers and then divide it by 2

Example:

1,2,3,4,5,6,7,8

The number 4 and 5 that is bolded will be divided

Example:

(4÷5)÷2

Hope it helps

7 0

3 years ago