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Vedmedyk [2.9K]
3 years ago
11

Find the total surface area of a cube whose edge is 2 . 1 m

Mathematics
1 answer:
galben [10]3 years ago
5 0

Answer:

Total surface of area =

6l {}^{2}

=6*2.1m

12.6m ^{2}

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What is the area, in square millimeters, of the parallelogram below?
mario62 [17]

Answer:

225

Step-by-step explanation:

If you move the bottom triangle up (as shown in the picture) it's not actually a square. And they haven't provided you with enough information to solve for anything unless there are still unknown values. You would need to know at least another leg length or angle. If they want the actual value in square millimeters, my best guess is that it's <em>supposed </em>to be a square. Since a square has sides of the same length, the area would be 15 * 15 which is 225.

3 0
3 years ago
What is the area of this trapezoid?
White raven [17]
The answer is 126in^2. The formula for calculating trapezoid is A=a+b/2(h). Area=base 1+base 2 divided by 2 times height. Base 1 is 12in. Base 2 is 2in+2in+12in=16in. And the height is 9in. So 12in+16in=28in. 28in/2=14in. 14in(9in)=126in^2.
**Note ^ before a number tells someone the number after it will be an exponent**
4 0
3 years ago
Write the number that is 1 less than 40,000
pickupchik [31]
40,000 minus 1 is 39,999
8 0
3 years ago
HELP!!
Liono4ka [1.6K]

Answer:

Part 1: The polygon ABCDE reflected across y-axis to get the polygon MNOPQ. So, the polygon ABCDE congruent to polygon MNOPQ.

Part 2: The vertices of polygon VWXYZ are V(1, 3), W(-1, 7), X(-7, 7), Y(-5, 3), and Z(-3, 1).

Step-by-step explanation:

Part 1:

The vertices of the polygon ABCDE are A(2, 8), B(4, 12), C(10, 12), D(8, 8), and E(6, 6).

The vertices of the polygon MNOPQ are M(-2, 8), N(-4, 12), O(-10, 12), P(-8, 8), and Q(-6, 6).

We need to find the transformation or sequence of transformations that can be performed on polygon ABCDE to show that it is congruent to polygon MNOPQ.

The relation between the vertices of ABCDE and MNOPQ are defined as

(x,y)\rightarrow (-x,y)

It means the polygon ABCDE reflected across y-axis to get the polygon MNOPQ. So, the polygon ABCDE congruent to polygon MNOPQ.

Part 2:

If polygon MNOPQ is translated 3 units right and 5 units down, then

(x,y)\rightarrow (x+3,y-5)

M(-2,8)\rightarrow V(-2+3,8-5)=V(1,3)

N(-4,12)\rightarrow W(-4+3,12-5)=W(-1,7)

O(-10,12)\rightarrow X(-10+3,12-5)=X(-7,7)

P(-8,8)\rightarrow Y(-8+3,8-5)=Y(-5,3)

Q(-6,6)\rightarrow Z(-6+3,6-5)=Z(-3,1)

Therefore the vertices of polygon VWXYZ are V(1, 3), W(-1, 7), X(-7, 7), Y(-5, 3), and Z(-3, 1).

6 0
3 years ago
2x+3y=0 and x+2y=-1 <br> Solve by substitution
dmitriy555 [2]

Answer:

The solution of the given equations is (-3,2)

Step-by-step explanation:

<u>Step:1</u>

The Given equations are 2 x+3 y=0 .......(1)

                                          x+2 y-1 =0.........(2)

<u>Step 2:-</u>

Solving<u> </u>both equations by using substitution method

multiply equation (2) with number '2' ,  

2 x+4 y-2=0.......(3)

subtracting the equations (1) and (3) and cancelled 'x' terms

we get 3y-4y=0-2

y=2

<u>Step 3:</u>-

substitute y= 2 in equation (1) we get

2 x+3 y=0

2 x+3(2)=0

2 x=-6

x=-3

<u>Final answer</u>:-

The solution of the given equations is (-3,2)

8 0
4 years ago
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