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k0ka [10]
3 years ago
12

If A = 10 feet, B = 6 feet, C = 1 foot, D = 14 feet, and E = 2 feet, what is the area of the object?

Mathematics
1 answer:
Bogdan [553]3 years ago
3 0

Answer:


Step-by-step explanation:


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The diagonal of a square is x units. What is the area of the square in terms of x?
bixtya [17]

Answer:

A = (x^2)/2

Step-by-step explanation:

Represent the length of one side of the square by s.  Then, according to the Pythagorean Theorem,

x^2 = s^2 + s^2, or x^2 = 2s^2, or (solving for s^2), s^2 = (x^2)/2

This last result is the area of the square in terms of x:  A = (x^2)/2

7 0
4 years ago
Find the mean of these values 6,4,8,2,5
AlekseyPX
6 + 4 + 8 + 2 + 5 = 25

25/5

mean = 5

5 0
4 years ago
HURRY PLEASE!! How many sides are there in a regular polygon which has 120 degrees in each of its angeles?
N76 [4]
Number of angles = number of sides

180 - 120 = 60 (calculating the exterior angle)

360 / 60 = 6

There are 6 sides.

Hope this helps! :)
7 0
3 years ago
Read 2 more answers
The sum of squares of three consecutive numbers is 77 find the numbers​
ollegr [7]

Answer:

4,5,6 are the three consecutive numbers. 16, 25 and 36 are their squares.

Step-by-step explanation:

Let the three consecutive numbers be x, (x+1), (x+2)

Now, the squares of these three numbers are  x^{2} ,(x+1)^{2} ,(x+2)^{2}

Sum = 77

∴by the problem ,

x^{2} +(x+1)^{2}+(x+2)^{2}  = 77\\x^{2} +(x^{2} +2x+1)+(x^{2}+4x +4) = 77\\x^{2} +x^{2} +2x+1+x^{2} +4x +4 = 77\\3x^{2} +6x+5 = 77\\3x^{2} +6x = 77-5\\3x^{2} + 6x = 72\\3x^{2} +6x-72= 0\\

{Taking 3 common }

x^{2} +2x- 24 = 0\\

{By factorization}

x^{2} +2x- 24 =0\\ x^{2} +6x-4x-24=0\\x(x+6)-4(x+6)=0\\(x+6)(x-4)=0\\

Therfore,

x= -6,4\\

<em>X can't be negetive </em>

∴ x =4\\x+1=5\\x+2=6

The squares of the three consecutive numbers are 16, 25, 36

The three consecutive numbers whose sum is 77 are 4, 5, 6

3 0
4 years ago
Find the value of x so that (-2, 4) is the midpoint between (x, 2) and (-5,3).
QveST [7]

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{x}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{-5}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{-5+x}{2}~~,~~\cfrac{3+2}{2} \right)~~=~~\stackrel{midpoint}{(-2,4)}\implies \begin{cases} \cfrac{-5+x}{2}=-2\\[1em] -5+x=-4\\ \boxed{x=1} \end{cases}

6 0
3 years ago
Read 2 more answers
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