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2 years ago

The legs of a right triangle have lengths of 4 and 5 units. what is the length of the hypotenuse

1 answer:

3 0

This question can be solved using the Pythagorean theorem. $a^2 + b^2 = c^2$ where a and b are the legs of the triangle and c is the hypotenuse.

$4^2 + 5^2 = c^2$
$16 + 25 = c^2$
$41 = c^2$
$\sqrt{41} = c$
So, c is approximately 6.403.

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lina2011 [118]

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3 years ago

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Which diagram contains a pair of similar triangles?

Musya8 [376]

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Step-by-step explanation:

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2 years ago

Find the equation of the line containing the points (0,3) and (1,8)

Tema [17]

Answer:

y = 5x + 3

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

To calculate m use the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (0, 3) and (x₂, y₂ ) = (1, 8)

m = $\frac{8-3}{1-0}$ = 5

Note the line crosses the y- axis at (0, 3) ⇒ c = 3

y = 5x + 3 ← equation in slope- intercept form

5 0

3 years ago

A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient

ankoles [38]

<u>Answer:</u>

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

<u>Solution:</u>

We need to show that the gradient of the curve at A is 1

Here given that ,

$y=(x-a) \sqrt{(x-b)}$ --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

$0=(b+1-a) \sqrt{(b+1-b)}$

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

$y^{\prime}=u^{\prime} y+y^{\prime} u$

so, we get

${u}^{\prime}=1$

$v^{\prime}=\frac{1}{2} \sqrt{(x-b)}$

$y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}$

By putting value of point A and putting value of eq 2 we get

$y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}$

$y^{\prime}=\frac{d y}{d x}=1$

Hence proved that the gradient of the curve at A is 1.

7 0

2 years ago

The altitude of an equilateral triangle is 9 inches. Find the perimeter of the triangle in inches.

Fiesta28 [93]

If we draw a perpendicular line from one of the vertices of the triangle we get 2 right angled triangles each with altitude 9 ins and vertex angle = 30 degrees. So:-

cos 30 = 9 /h where h = one of the sides of the equilateral triangle

h = 9 / cos 30 = 10.392 inches

Therefore the perimeter of the triangle = 3 * 10.392 = 31.1769 ins

Answer is 31.18 inches to the nearest hundredth.

5 0

2 years ago