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3 years ago

In a right triangle the lengths of the legs are a and b. Find the length of a hypotenuse, if: a=3, b=4;

Mathematics

2 answers:

prohojiy [21]3 years ago

6 0

Answer:

The hypotenuse is 5

Step-by-step explanation:

We can use the Pythagorean theorem to find the hypotenuse of a right triangle

a^2 +b^2 =c^2 where a and b are the legs and c is the hypotenuse

3^2 +4^2 =c^2

9+16=c^2

25=c^2

Take the square root of each side

sqrt(25) =sqrt(c^2)

5 = c

Lady bird [3.3K]3 years ago

4 0

Answer:

<h3>hypotenuse = 5</h3>

Step-by-step explanation:

Use the Pythagorean theorem:

a² + b² = c²

a, b - legs

c - hypotenuse

We have the lengths of the legs a = 3 and b = 4. Substitute:

c² = 3² + 4²

c² = 9 + 16

c² = 25 → c = √25

c = 5

Pythagorean triple:

3, 4, 5

6, 8, 10

5, 12, 13

and more.

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If x / 2y = 0.4 , A / 5B = 0.25 Then A + y / 5 =…....

shusha [124]

Answer:

$\frac{x}{2y} = \frac{4}{10} \\ 8y = 10x \\ y = \frac{10x}{8} - - - - - 1 \\ \\ \frac{a}{5b} = \frac{25}{100} \\ \frac{a}{5b} = \frac{1}{4} \\ 4a = 5b \\ a = \frac{5b}{4} - - - 2 \\ \\ so \: \: \frac{a + y}{5} \\ = (a + y) \times \frac{1}{5} \\ = ( \frac{5b}{4} + \frac{5x}{4} ) \times \frac{1}{5} \\ = \frac{5b + 5x}{4} \times \frac{1}{5} \\ = \frac{5(b + x)}{4} \times \frac{1}{5} \\ = \frac{b + x}{4} \\ \\ so \: \: \: \frac{a + y}{5} = \frac{b + x}{4}$

6 0

3 years ago

What is the area of the figure below a. 7.5 b. 15 c. 21.25 d. 42.5

Marina86 [1]

Check the picture below.

let's recall that in a Kite, the diagonals meet at 90° angles, therefore, we know the height of each of those 4 triangles, is 2.5 and 6, now, since the pair of triangles above are 45-45-90 triangles, we can use the 45-45-90 rule, as you see there, so, if the height is 2.5, then the base is also 2.5.

so, we really have 2 pair of triangles whose base is 2.5 and height of 2.5, and another pair of triangles whose base is 2.5 and height is 6, let's add their areas.

$\bf \stackrel{\textit{area of 2 triangles above}}{2\left[\cfrac{1}{2}(2.5)(2.5) \right]}~~+~~\stackrel{\textit{area of 2 triangles below}}{2\left[ \cfrac{1}{2}(2.5)(6) \right]}\implies 6.25+15\implies 21.25$

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3 years ago

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Lorenzo drew 3 triangles, 4 rectangles, and 7 arrows in a set of 20 shapes. The only other type of shape in the set is a circle.

djyliett [7]

Answer:

$\large\boxed{6/20}$

Step-by-step explanation:

In this question, we're trying to find how many circles are there and represent it as a fraction.

We know that he drew:

3 triangles
4 rectangles
7 arrows
Set of 20 shapes

We need to find how many circles are there.

3 tiangles + 4 rectangles + 7 arrows = 14 shapes

We know that there is suppose to be 20 shapes, this means that the remaining amount would be circles (6 circles)

We now know that there are 6 circles and 20 shapes in total.

This means that the fraction that would represent circles is 6/20

<h3>I hope this helped you out.</h3><h3>Good luck on your academics.</h3><h3>Have a fantastic day!</h3>

8 0

3 years ago

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X х<br> 6<br> 16<br> +<br> =<br> x - 4<br> x - 3<br> x2 – 7x + 12

Inessa [10]

Answer:

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3 years ago

f(x)=6x3−54x2−126x−8 is decreasing on the interval ( equation editorEquation Editor , equation editorEquation Editor ). It is in

worty [1.4K]

Answer:

The function $f(x)=6x^3-54x^2-126x-8$ is decreasing on the interval $(-1,7)$ and it is increasing on the interval $(-\infty, -1)\cup (7, \infty)$

Step-by-step explanation:

To determine the intervals of increase and decrease of the function $f(x)=6x^3-54x^2-126x-8$ , perform the following steps:

1. Differentiate the function

$\frac{d}{dx}\left(6x^3-54x^2-126x-8\right)=\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\\frac{d}{dx}\left(6x^3\right)-\frac{d}{dx}\left(54x^2\right)-\frac{d}{dx}\left(126x\right)-\frac{d}{dx}\left(8\right)\\\\f'(x)=18x^2-108x-126$

2. Obtain the roots of the derivative, f'(x) = 0

$\mathrm{Factor\:out\:common\:term\:}18:\quad 18\left(x^2-6x-7\right)\\\\\mathrm{Factor}\:x^2-6x-7:\quad \left(x+1\right)\left(x-7\right)\\\\18x^2-108x-126:\quad 18\left(x+1\right)\left(x-7\right)\\\\18x^2-108x-126=0\quad :\quad x=-1,\:x=7$