All categories

3 years ago

What is the approximate length of the diagonal of a square with side length of 20 centimeters

Mathematics

2 answers:

zhuklara [117]3 years ago

5 0

Answer:28.28 cm

Explanation:

a² + b² = c²

20² + 20² = c²

c² = 400 + 400

c² = 800

c = √800

c = 28.28 cm

Jet001 [13]3 years ago

4 0

You can either use the Pythagorean theorem or the formula d = a√2.

Pythagorean theorem:

The diagonal is the hypotenuse of the right triangle in the square and the two sides are the legs.

20² + 20² = d²
400 + 400 = d²
800 = d²
√800 = d
28.28 = d

Using the formula:

d = a√2
d = 20√2
d = 28.28

The answer with both the methods is the same.

Hope this helps :)

You might be interested in

A textbook costs $65.49 before tax. The tax on the textbook is 6.5%. What is the total cost of the textbook? Round your answer t

AnnZ [28]

65.49*0.065 = 4.26
65.49+4.26 = $69.75

7 0

2 years ago

Read 2 more answers

What is the value of x in the equation 8x+16

nevsk [136]

Answer:2

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

PLEASE HELP ME QUICK!!!20 POINTS!!!!!!!!

Blizzard [7]

the volume scales by (1/10)^3. so it would be 48.875*10^-3.

5 0

3 years ago

WILL MARK CORRECT ANSWER BRAINLIEST

vladimir1956 [14]

3y^-4/3 x 2 3/y
Write in exponential form with the base of y
= 3y^-4/3 x 2y^1/3
Calculate the product
= 6y^-1
Any expression raised to the power of equals its reciprocal
= 6 x 1/y
Calculate the product
= 6/y

8 0

2 years ago

Find the equation for the plane that contains the line x=−1+3t , y=1+2t, z=2+4t and is perpendicular to the plane containing the

Ivan

Let $L$ be the line given by the vector equation

$(-1,1,2) + \lambda(3,2,4) \ , \lambda \in \mathbb{R}$ .

First, we use the director vectors of the lines L1 and L2 to get the

vector equation of the plane containing them, which we denote by $\Pi_1$ . This is,

$\\\\\Pi_1 : (1,-1,5) + \alpha (2,-1,6) + \beta (1,1,-3) \ , \alpha, \beta \in \mathbb{R}\\\\\\$

We observe that $\vec{N} = (2,-1,6)\times(1,1,-3) = (-3,12,3) \ne (0,0,0)$ . Therefore, the vector equation of $\Pi_1$ defines a plane and $\vec{N}$ is a normal vector to $\Pi_1.$

Finally, the vector equation for the wanted plane, which we denote by $\Pi$ , is

$\Pi : (-1,1,2) + r(3,2,4) + s(-3,12,3), r,s \in \mathbb{R} \ .$

Thus, if $s = 0$ , then $L \subset \Pi$ and since $\vec{N}$ is parallel to $\Pi$ , then it is perpendicular to $\Pi_1$ .

8 0

2 years ago