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Zarrin [17]
3 years ago
5

Two positive integers have a product of 200.

Mathematics
1 answer:
Alexxx [7]3 years ago
6 0

Answer:

x=10

the two integers are 10, 20

Step-by-step explanation:

x=integer 1

2x=integer 2

2x(x)=200

2x^{2}=200

divide both sides by 2

x^{2}=100

take the square root

√x^{2}=√100

x=10

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Write equations for the vertical and horizontal lines passing through the point , −1−8 .
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Vertical line (when y = 0)
⇒ Equation : x = -1

Horizontal line (when x = 0)
⇒ Equation : y = -8

--------------------------------------------------------------
Answer: x = -1, y = -8
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7 0
4 years ago
A man sells a sofa for rs 3210 making a profit of 7% what would have been his profit percent if he had sold it for rs 3360?
Digiron [165]
Let x be the original cost for the sofa.
X x (1 + 0.07) =3210
X = 3000
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There is a bag filled with 5 blue, 6 red and 2 green marbles.
nika2105 [10]

Answer:

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8 0
3 years ago
Help me with this urgent due in 5 min​
solniwko [45]

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The last option

Step-by-step explanation:

4 0
3 years ago
<img src="https://tex.z-dn.net/?f=The%20%5C%3A%20%20value%20%20%5C%3A%20of%20%20%5C%3A%20%20%5Csqrt%7B6%20%2B%20%20%5Csqrt%7B6%2
saul85 [17]

\large\underline{\sf{Solution-}}

<u>Let us assume that:</u>

\sf \longmapsto x =  \sqrt{6 +  \sqrt{6 +  \sqrt{6 + ... \infty } } }

We can also write it as:

\sf \longmapsto x =  \sqrt{6 +  x }

Squaring both sides, we get:

\sf \longmapsto  {x}^{2}  =6 +  x

\sf \longmapsto  {x}^{2} - x - 6  =0

By splitting the middle term:

\sf \longmapsto  {x}^{2} - 3x + 2x - 6  =0

\sf \longmapsto x(x - 3) + 2(x - 3 ) =0

\sf \longmapsto (x+ 2)(x - 3 ) =0

<u>Therefore:</u>

\longmapsto\begin{cases} \sf (x+ 2) =0 \\ \sf (x - 3) = 0 \end{cases}

\sf \longmapsto x =  - 2,3

<u>But x cannot be negative. </u>

\sf \longmapsto x = 3

Therefore, the value of the expression is 3.

\large\underline{\sf{Verification-}}

Given:

\sf\longmapsto x=3

We can also write it as:

\sf\longmapsto x = \sqrt{9}

\sf\longmapsto x = \sqrt{6+3}

\sf\longmapsto x = \sqrt{6 + \sqrt{9}}

\sf\longmapsto x = \sqrt{6 + \sqrt{6+3}}

\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{9}}}

\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+3}}}

This pattern will continue.

\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+...\infty}}}

7 0
3 years ago
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