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pashok25 [27]
3 years ago
15

A square has sides of length 70 meters. What is the perimeter?

Mathematics
1 answer:
soldier1979 [14.2K]3 years ago
6 0

Answer:

280

Step-by-step explanation:

If a square has sides of length 70 meters, the perimeter is 280.

Formula: P=4a

P = 4a = 4·70 = 280

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4 2/3 divided by 1/2
ycow [4]

Answer:

9 1/3

Step-by-step explanation:

1. Make 4 2/3 into a improper fraction

4 2/3= 14/3

2. Keep, change, flip

14/3 x 2/1

3. Multiply

14/3 x 2/1= 28/3

4. Simplify

28/3= 9 1/3

(btw keep, change, flip basically means to keep the first fraction the way it is, change the division sign to a multiplication sign, then flip the second fraction so the original denominator is the new numerator, and vise versa.)

3 0
3 years ago
Read 2 more answers
Use euler's formula to find the missing number <br><br>vertices: 11<br>edges: 34<br>Faces: ?
larisa [96]

Answer:

F = 25

Step-by-step explanation:

Euler's formula for polyhedra states that

F + V - E = 2

Here V = 11, E = 34, thus

F + 11 - 34 = 2

F - 23 = 2  ( add 23 to both sides )

F = 25

8 0
3 years ago
Which shape is both a parallelogoram and a rhombus
Morgarella [4.7K]

Answer:

A square can be defined as a rhombus which is also a rectangle and a parellelogram with four congruent sides and four right angles

I hope this helps

3 0
3 years ago
What is 3 1/3 as a percent?
Nikitich [7]

Answer:

333.3\%

Step-by-step explanation:

We want to express 3⅓ as a percentage.

Let us first of all rewrite 3⅓ as decimals to obtain:

3 \frac{1}{3}  = 3.3333...

To express this as a percentage, we multiply by 100%

3 \frac{1}{3}  = 3.3333... \times 100\%

This multiplies to give us:

3 \frac{1}{3}  = 333.3\%

4 0
3 years ago
The difference of two positive numbers is 7 and the sum of their squares is 109. Find the numbers.
Ray Of Light [21]

Answer:

Numbers\ are\ 58\ and\ 51.

Step-by-step explanation:

The the numbers are x\ and\ y.

<em>Difference:</em>

<em></em>Difference\ of\ numbers=7\\\\x-y=7....................eq(1)<em></em>

<em>Sum:</em>

<em></em>Sum\ of\ numbers=109\\\\x+y=109...........................eq(2)<em></em>

<em></em>

Solve eq(1) and eq(2).

eq(1)+eq(2)\\\\x-y+x+y=7+109\\\\2x=116\\\\x=58\\\\Now\ from\ equation\ 1\\\\x-y=7\\\\y=x-7\\\\y=58-7\\\\y=51

7 0
3 years ago
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