All categories

2 years ago

Find the missing length for a right triangle with hypotenuse of 18 and a leg length of 6

Mathematics

2 answers:

Alla [95]2 years ago

8 0

16.9705624748447714
if you want it rounded its 16.97 or 16.9
thanks for letting me answer and god bless <3

yaroslaw [1]2 years ago

4 0

$\sqrt{18^2-6^2}$
$\sqrt{256-36} = \sqrt{220} = 2 \sqrt{55}$

You might be interested in

Please answer this correctly

Hoochie [10]

Answer:

3:48 P.M

because you add all the number making sure it's not greater than 60 and carry over to the next hour if it is

3 0

2 years ago

Read 2 more answers

You are planning to make an open rectangular box from aa 4141-in.-by-8181-in. piece of cardboard by cutting congruent squares

marissa [1.9K]

Answer:

13058.83 cubic inches

Step-by-step explanation:

Given that a rectangular box is having dimensions as 41x81 inches.

Let x be the side of square cut from all the four corners.

The open box made would have height as x and length 41-2x with width 81-2x

Volume =

$V(x) = x(41-2x)(81-2x)\\V(x) =x(3321-244x+4x^2)\\V(x) = 3321x-244x^2+4x^3\\V'(x) = 3321-488x+12x^2\\V"(x) = -488+24x\\$

Equate first derivative to 0

We get applicable root as x = 8.642

Max volume = 13058.83 cubic inches

3 0

2 years ago

overnight a pipe cracked and started to leak into a basement. the graph shows the depth of the water in relation to time. how ma

Shkiper50 [21]

Answer:

if a water pipe bursts over night depends on the pressure of the water but i will say a foot or more an hour

Step-by-step explanation:

3 0

2 years ago

What is the volume of the figure?

Fynjy0 [20]

Volume of Cylinder

$V = \pi r^2 h$

$V = \pi \times (11~yd)^2 \times 9~yd$

$V = 1089 \pi~yd^3$

$V = 3421~yd^3$

3 0

2 years ago

Find the coordinates for the midpoint of the segment with endpoints give (3,5) and (-2,0)

Leto [7]

Answer:

The coordinates of the mid-point are :

$(\frac{1}{2}, \frac{5}{2})$

Step-by-step explanation:

We know that, the coordinates of the mid-point (x, y) of a line segment joining the points (x₁, y₁) and (x₂, y₂) is given by

$(x, y) = (\frac{x_{1} +x_{2} }{2},\frac{y_{1}+y_{2} }{2} )$

Now, we have the given points as (3, 5) and (-2, 0).

By using the above formula, coordinates of the mid-point (x, y) of the line-segment joining the points (3, 5) and (-2, 0) is given by,

$x = \frac{3+(-2)}{2} = \frac{3-2}{2} = \frac{1}{2}$

$y = \frac{5+0}{2} = \frac{5}{2}$

∴ coordinates of the mid-point of the line-segment joining the points (3,5) and (-2,0) is $(\frac{1}{2}, \frac{5}{2})$ .

8 0

2 years ago