All categories

3 years ago

How to find area of a shape

1 answer:

7 0

Answer: you would tend to take the main dimensions of the shape like height times width and add what you need for the shape that you are finding the area for ; circles= pie2r trinagle=h*w/2 squre=s2 rectangle w*h

Step-by-step explanation:

You might be interested in

The price of a stock had a change of –3 dollars.

Leokris [45]

The absolute value of -3 is -3

5 0

3 years ago

Read 2 more answers

on day 1 a patient measures 5 feet and 11 inches tall. after 4 days of stretching the patient measures 5 feet and 11 inches tall

Vinvika [58]

Answer:

slope is zero

Step-by-step explanation:

zero slope is neither positive or negative; it indicates no change whatsoever

5 0

3 years ago

A group of friends are going to a cottage for the weekend and you look at the road map to see how far away it is.

ioda

8.5 x 250,000 = 2,125,000
2,125,000 x .00001 = 21.25km
1cm = .00001km

7 0

3 years ago

Solve by using Pythagorean Theorem, a= 5, b= x-1, and c=x

charle [14.2K]

The value of x is 13, from the values of x the values of a,b and c becomes 5, 12 and 13, which is obtained by using Pythagorean theorem.

Step-by-step explanation:

The given is,

a = 5

b = x - 1

c = x

Step:1

Pythagorean theorem is,

$a^{2} + b^{2} = c^{2}$ ................................(1)

( It states that the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares on the other two sides )

From the given values the equation (1) becomes,

$5^{2} + (x-1) ^{2} = x^{2}$

( $(a-b)^{2} = a^{2} + b^{2} -2ab$ )

25 + ( $x^{2}$ + 1 - 2x ) = $x^{2}$

25 + 1 -2x = 0 ( ∵ Eliminate $x^{2}$ in both sides)

26 = 2x

x = 13

Step:2

From the values of x,

a = 5

b = ( 13 - 1 ) = 12

c = 13

Result:

The value of x is 13, from the values of x the values of a,b and c becomes 5, 12 and 13, which is obtained by using Pythagorean theorem.

3 0

3 years ago

Which could be the first step and simplifying this expression? Check all that apply.

Triss [41]

$\bf \left( x^3x^{-6} \right)^2\implies \left( x^{3-6} \right)^2\implies \stackrel{1st~Step}{\left( x^{-3} \right)^2}\implies x^{-6}\implies \cfrac{1}{x^6}\\ \qquad \Downarrow\\ \stackrel{\textit{distributing the exponent}}{\left( x^{3\cdot 2}x^{-6\cdot 2} \right)}\\ \qquad \Downarrow\\ \stackrel{1st~Step}{x^6x^{-12}}\\ \qquad \Downarrow\\ x^{6-12}\\ \qquad \Downarrow\\ x^{-6}\\ \qquad \Downarrow\\ \cfrac{1}{x^6}$

7 0

3 years ago