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castortr0y [4]
3 years ago
15

Find two positive consecutive odd integers whose products is 99.

Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
5 0

Answer:

9,11

Step-by-step explanation:

let the numbers be x, x+2 (since consecutive odd integers have a difference of 2)

therefore;

x(x+2)= 99

x²+2x=99

I usually solve the sum like this,

we need to find the nearest square number ( number smaller than the number on the R.H.S)

here, it is 81, whose square root is 9. therefore the solution is 9

verification:

(9)²+2*9=99

81+18=99

99=99

hence, 9 is the solution. therefore the positive odd consecutive integers are  

x=9

x+2=9+2= 11

verification:

9*11=99

99=99

L.H.S=R.H.S

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Find parametric equations for the line through point P(-4,5,2) in the direction of the vector (-6,6,-5)
tigry1 [53]

The vector <em>v</em> (-6, 6, -5) points in the direction of itself, so we start there.

We can capture all points on the line through the origin and the point (-6, 6, -5) by scaling <em>v</em> by an arbitrary real number <em>t</em>.

The line through point <em>P</em> and pointing in the same direction as <em>v</em> is parallel to the other line that passes through the origin. Then the line we want can be obtained by translating the line through the origin by a vector <em>p</em> that points to (-4, 5, 2), so the vector equation for this line is

<em>r </em>(<em>t </em>) = <em>p</em> + <em>t v</em>

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<em>r </em>(<em>t </em>) = (-4 - 6<em>t</em>, 5 + 6<em>t</em>, 2 - 5<em>t</em> )

To get the parametric equations, simply take out the components:

<em>x</em> (<em>t</em> ) = -4 - 6<em>t</em>

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<em>z</em> (<em>t</em> ) = 2 - 5<em>t</em>

4 0
3 years ago
Please help me on this i don’t get it that well
NeX [460]

Answer:

13.75

Step-by-step explanation:

make a ratio between the original lengths, 11 : 5.5

then put the ratio of the enlarged lengths, 27.5 : __

now you divide 27.5 by 11 to find how many times you multiplied the lengths, 2.5.

now you multiply 5.5 by 2.5 giving you the answer of 13.75

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3 years ago
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8 0
3 years ago
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