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REY [17]
4 years ago
10

What is the sum of all the integers between the square root of 10 and the square root of 47

Mathematics
2 answers:
polet [3.4K]4 years ago
8 0

What's the square root of  10 ?
     3²=9  and  4²=16
     So  √10  is somewhere between  3  and  4 .

What's the square root of  47 ?
     6²=36  and  7²=49
     So  √47  is somewhere between 6 and 7 .

The integers between √10 and √47 are  4,  5, and  6 .

Their sum is (4 + 5 + 6) = 15 .

insens350 [35]4 years ago
5 0
First estimate the square root of 10, about 3.16
and then the square root of 47, about 6.85
now what are the integers in between 3.16 and 6.85:
4,5,6
now add them all 
4+5+6
=15
Hope this helps.
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(1 point) find the area of the triangle in the plane with vertices (1,4)(1,4), (5,5)(5,5), and (−2,9)(−2,9)
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4 0
4 years ago
<img src="https://tex.z-dn.net/?f=-6%20%2B%203x%20-%209x%5E%7B2%7D%3D-16" id="TexFormula1" title="-6 + 3x - 9x^{2}=-16" alt="-6
ehidna [41]

The solution of x in -6 + 3x - 9x^2 = -16 is x = \frac{3 \pm \sqrt{369}}{18}

<h3>How to solve the equation?</h3>

The equation is given as:

-6 + 3x - 9x^2 = -16

Add 16 to both sides

10 + 3x - 9x^2 = 0

Rewrite as:

9x^2 - 3x - 10 = 0

Solve for x using the quadratic formula

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

So, we have:

x = \frac{3 \pm \sqrt{(-3)^2 - 4 * 9 * -10}}{2 * 9}

Evaluate

x = \frac{3 \pm \sqrt{369}}{18}

Hence, the solution of x in -6 + 3x - 9x^2 = -16 is x = \frac{3 \pm \sqrt{369}}{18}

Read more about quadratic functions at:

brainly.com/question/1497716

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4 0
2 years ago
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11)y=-2/3x+8
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3 0
3 years ago
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finlep [7]
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b = the point that intercepts the y axis; the y-intercept
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