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avanturin [10]
3 years ago
8

What is the sum of 36 plus 25

Mathematics
2 answers:
Ede4ka [16]3 years ago
6 0
36 + 25 = 30 + 20 = 50
6 + 5 = 11
50 + 11 = 61
61 is the answer
Hope this helps!
bija089 [108]3 years ago
5 0
The sum is 61 for this equation.
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bekas [8.4K]

Mutually exclusive means two things that cannot happen simultaneously. John cannot eat 4 apples at a time. I hope that helps:)

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3 years ago
A set of numbers is shown below:
eimsori [14]

Answer:

<em><u>[</u></em><em><u>2</u></em><em><u>,</u></em><em><u>4</u></em><em><u>,</u></em><em><u>6</u></em><em><u>]</u></em>

Step-by-step explanation:

GIVEN EQUATION : 2X+3≥7

i.e, 2x≥7-3

i.e, 2x≥4

i.e, x≥4/2

i.e, x≥2

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3 years ago
How do you write out in expanded form 35.245 in fractions
Doss [256]

I believe the answer to this30+5+.2+.04+.005

3 0
2 years ago
Read 2 more answers
One of the roots of the quadratic equation x2−5mx+6m2=0 is 36. Find the greatest possible value of the second root. Help needed
Ivenika [448]
<h2>The greatest possible value of the second root, β = 54 </h2>

Step-by-step explanation:

The given quadratic equation:

x^2-5mx+6m^2=0

Let α and β be the roots of the given quadratic equation.

α = 36

To find,  the greatest possible value of the second root ( β) = ?

∴ The sum of the roots,

α + β = \dfrac{-b}{a}

⇒ 36 + β = \dfrac{-(-5m)}{1}

⇒ 5m = 36 + β               ............. (1)

The product of the roots,

α.β = \dfrac{c}{a}

⇒ 36.\beta=\dfrac{6m^2}{1}

⇒ 6.\beta=m^2                   ............. (2)        

From equations (1) and (2), we get  

⇒ (\dfrac{36+\beta}{5})^{2}=6\beta

⇒ \beta^2-78\beta+1296=0

⇒ \beta^2-54B\beta-24B\beta+1296=0

⇒ β(β - 54) - 24(β - 54) = 0

⇒ (β - 54)(β - 24) = 0

⇒ β - 54 = 0 or, β - 24 = 0

⇒ β = 54 or, β = 24

∴  The greatest possible value of the second root, β = 54

4 0
3 years ago
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kifflom [539]
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ok, let's say the shortest is "a"
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and let's say "c" is -> "</span><span>3rd side 14 more feet than length of shortest side"
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we know the perimeter ( sum of all sides ) is 154

</span>\bf \begin{array}{llll}&#10;a&b&c\\&#10;&\downarrow&\downarrow  \\&#10;&2a&a+14\\&#10;\downarrow &\downarrow &\downarrow \\&#10;\end{array}\\&#10;\ a+b+\ c=154\\\\&#10;a+(2a)+(a+14)=154
<span>
solve for "a"</span>
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3 years ago
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