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svetlana [45]
4 years ago
6

The perimeter of a square measures 1.3 meters. What is its side length in centimeters?

Mathematics
1 answer:
ololo11 [35]4 years ago
5 0

Side length in centimeters is

A) 130 cm

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8X - 35 = 7(1 + 2X)​
lawyer [7]

Answer: x=-7

Step-by-step explanation:

8X - 35 = 7(1 + 2X)​

8x - 35 = 7 + 14x

-6x - 35 = 7

-6x = 42

X = -7

3 0
3 years ago
If x^2=20 what is the value of x will give brainliest for answer
Schach [20]

Answer:

x² - 20 = 0

Using the quadratic formula

x =  \frac{ - b± \sqrt{( {b})^{2} - 4ac } }{2a}

a = 1 b = 0 c = -20

So we have

x =  \frac{ - 0 ±  \sqrt{ {0}^{2}  - 4(1)( -20)} }{2(1)}  \\  \\ x =  \frac{± \sqrt{80} }{2}  \\  \\ x =  \frac{±4 \sqrt{5} }{2}  \\  \\  \\ x = ±2 \sqrt{5}  \\  \\  \\ x = 2 \sqrt{5}  \:  \:  \: or \:  \:  \: x =  - 2 \sqrt{5}

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8 0
3 years ago
How many ounces of a silver alloy that costs $5.50 per ounce should be mixed with one that costs $7.00 per ounce to make a new 3
Pepsi [2]

Answer:

<u>135.73 ounces</u> of a silver alloy that costs $5.50 per ounce should be mixed.

Step-by-step explanation:

Given:

A silver alloy that costs $5.50 per ounce should be mixed with one that costs $7.00 per ounce to make a new 30 ounce alloy that costs $6.40 per ounce.

Now, to find the ounces of silver alloy.

Let the silver costs $5.50 per ounce be x.

And the silver costs $7.00 per ounce be y.

So, the total ounce make a new alloy:

x+y=30\\y=30-x  ....(1)

Now, the total costs of silver alloy:

5.50x+7y=6.4

Putting the value of y from equation (1) in the place of y :

5.50x+7(30-x)=6.4

5.50x+210-7x=6.4

-1.5x+210=6.4

<em>Subtracting both sides by 210 we get:</em>

-1.5x=-203.6

<em>Dividing both sides by -1.5 we get:</em>

x=135.73

Therefore, 135.73 ounces of a silver alloy that costs $5.50 per ounce should be mixed.

7 0
3 years ago
For cones with radius 6 units, the equation V = 127th relates the height h of the cone, in units, and the volume V of the cone,
denis-greek [22]

Answer:

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Step-by-step explanation:

cómo estás yo bien compita

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3 years ago
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