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2 years ago

Find the missing right triangle side lengths in the models of the Pythagorean Theorem below

Mathematics

2 answers:

Sladkaya [172]2 years ago

8 0

Answer:

The picture on the left doesn’t have a right triangle.

maw [93]2 years ago

5 0

Answer: 24, 10, 26 and 4.6, 5, 6.7

Step-by-step explanation: for the first triangle the square with the area 576 has side lengths of 24 because 24 is the square root of 576, the square with the area 676 has a side length of 26 because 26 is the square root of 576, and we already know the side lengths of the third triangle are 10 for the second triange we know the side length of the square with the area 21 is about 4.6 because √21 ≈ 4.6 and the square with the area 25 the side length is 5 because √25 = 5 and the last square the side length is 6.7 because √46 ≈ 6.7

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Citrus2011 [14]

The answer is -15 yayaya

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2 years ago

−6 + 2n = 3n − ( 6 + 5 ) Please provide the answer along with how you got the answer!

pickupchik [31]

Answer:

Step-by-step explanation:

-6+2n=3n-(6+5)

-6+2n=3n-11

-6+2n-3n=-11

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2 years ago

Given that f(x) = 2x – 1, find f(6). Your answer: 6 7 11 13

Kruka [31]

Answer:

Step-by-step explanation:

2 times 6 is 12 minus 1 is 11

4 0

2 years ago

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2. Given a quadrilateral with vertices (−1, 3), (1, 5), (5, 1), and (3,−1):

zlopas [31]

<h2>Explanation:</h2>

In every rectangle, the two diagonals have the same length. If a quadrilateral's diagonals have the same length, that doesn't mean it has to be a rectangle, but if a parallelogram's diagonals have the same length, then it's definitely a rectangle.

So first of all, let's prove this is a parallelogram. The basic definition of a parallelogram is that it is a quadrilateral where both pairs of opposite sides are parallel.

So let's name the vertices as:

$A(-1,3) \\ \\ B(1,5) \\ \\ C(5,1) \\ \\ D(3,-1)$

First pair of opposite sides:

Slope:

$\text{For AB}: \\ \\ m=\frac{5-3}{1-(-1)}=1 \\ \\ \\ \text{For CD}: \\ \\ m=\frac{1-(-1)}{5-3}=1 \\ \\ \\ \text{So AB and CD are parallel}$

Second pair of opposite sides:

Slope:

$\text{For BC}: \\ \\ m=\frac{1-5}{5-1}=-1 \\ \\ \\ \text{For AD}: \\ \\ m=\frac{-1-3}{3-(-1)}=-1 \\ \\ \\ \text{So BC and AD are parallel}$

So in fact this is a parallelogram. The other thing we need to prove is that the diagonals measure the same. Using distance formula:

$d=\sqrt{(y_{2}-y_{1})^2+(x_{2}-x_{1})^2} \\ \\ \\ Diagonal \ BD: \\ \\ d=\sqrt{(5-(-1))^2+(1-3)^2}=2\sqrt{10} \\ \\ \\ Diagonal \ AC: \\ \\ d=\sqrt{(3-1)^2+(-5-1)^2}=2\sqrt{10} \\ \\ \\$

So the diagonals measure the same, therefore this is a rectangle.

5 0

3 years ago

Find the product. (-5a 2)3 · a 5 -15a10 -125a11 15a6 -125a10

alekssr [168]

The correct answer is not up there. its -125a^7

3 0

3 years ago

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