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torisob [31]
2 years ago
6

Help me! i will give brainilest to thnnswer!!!!!!!!!!1e correct a

Mathematics
1 answer:
o-na [289]2 years ago
6 0
Because the square root of 36 is 6, so if the number on question A is 37 then the answer would be larger than 6. The same goes for the other questions. The square root of 100 is 10 so since the number is 95 is with be smaller than 10. And for the last one the square root of 25 is 5 and the square root of 36 is 6, So the square root of 30 is in between 5 and 6. Hope this helps!!
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Seven tenths of a number plus 14 is less than forty-nine
Virty [35]
7/10n + 14 = 49
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n = 35 x 10/7
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What brand is better ? Brand A: $2.50 Brand B: 35 for $3.25 Brand C: 50 for $5.25 Brand D: 100 for $10.00
arlik [135]

I belive that the answer should be D) 100 for $10.00 because I had this exact same question one time.

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3 0
2 years ago
Quadrilateral ABCD has the following vertics:A( -3,0) B (7,2) C (1,-4) D (-9,-6) is Quadrilateral A B C D a parallelgram, and wh
alekssr [168]

Answer:

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3 years ago
Given an exponential function for compounding interest A(x)= P(.82)^x what is the rate of change
Katena32 [7]

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3 0
2 years ago
Which is the equation of a hyperbola centered at the origin with x-intercept +\- 3 and asymptote y=2x
Radda [10]

Answer:

{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}

Step-by-step explanation:

The hyperbola has x-intercepts, so it has a horizontal transverse axis.

The standard form of the equation of a hyperbola with a horizontal transverse axis is  \dfrac{(x - h)^{2}}{a^{2}} - \dfrac{(y - k)^{2}}{b^{2}} = 1

The center is at (h,k).

The distance between the vertices is 2a.

The equations of the asymptotes arey = k \pm \dfrac{b}{a}(x - h)

1. Calculate h and k. The hyperbola is symmetric about the origin, so  

h = 0 and k = 0

2. For 'a': 2a = x₂ - x₁ = 3 - (-3) = 3 + 3 = 6

a = 6/2 = 3  

3. For 'b': The equation for the asymptote with the positive slope is  

y = k + \dfrac{b}{a}(x - h) = \dfrac{b}{a}x

Thus,  asymptote has the slope of

\begin{array}{rcl}m& =& \dfrac{b}{a}\\\\2& =& \dfrac{b}{3}\\\\b& =& \mathbf{6}\end{array}

4.  The equation of the hyperbola is

\large \boxed{\mathbf{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}}

The attachment below represents your hyperbola with x-intercepts at ±3 and asymptotes with slope ±2.

7 0
2 years ago
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