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4 years ago

Need to turn it in soon! help!

Mathematics

1 answer:

olchik [2.2K]4 years ago

5 0

5: is false because A rectangle is a four-sided shape where every angle is a right angle (90°).Also opposite sides are parallel and of equal length.

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Miki has a sheet of paper with a length of 104 centimeters and a width of 88 centimeters. She wants to divide it into the larges

LuckyWell [14K]

First, find the area of the rectangular paper: 104 × 88 = 9152 cm²

Since we want to find the number of the largest square that we can cut from the paper, we need to investigate if there's any square number is the factor of 9152

We can use the prime factor tree to find the factors of 9152 that is also a square number as shown below

9152 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 13

after trying few combinations, we have:
2 × 2 × 2 × 2 = 16
2 × 2 × 2 × 2 × 2 × 2 = 64

The largest square number is 64, and we will have 9152 ÷ 64 = 143 squares of papers

4 0

4 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 $s_{ABD} = \frac{AB+BD+AD}{2}$