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kaheart [24]
3 years ago
11

Does the equation x^2 -4x + y^2 = -3 intersect the x-axis?

Mathematics
2 answers:
frutty [35]3 years ago
5 0
We can rewrite the equation into:
(x-2)^2+y^2=1

Therefore the circle's centre is that (2,0). Since the circle's centre lies on the x axis, the circle must intersect the x-axis.
olya-2409 [2.1K]3 years ago
3 0

Answer:

Yes, because the center is on the x-axis?

Step-by-step explanation:

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Find an equation for a tangent plane z=e-x2-y2 at the point (0,0,1).
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Step-by-step explanation:

Given equation

z = e^(-x²-y²) at point (0,0,1)

now let z = f(x,y)

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now

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equation of the tangent plane therefore will be

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Which of the following numbers can be expressed as a decimal that terminates? 3/2, 2/3, 3/4, 5/7
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Answer:

3/2 and 3/4 are terminating decimals

Step-by-step explanation:

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Answer:

<h2><u>question 1</u></h2>

n^{2} - 20n -96 = 0

use product and sum method

product = -96

sum = -20

numbers needed = ( -24 , 4)

n - 24 = 0

n + 4 = 0

hence <u>n = 24 and n = -4 </u>

<u></u>

<h2><u>Question 2 </u></h2>

<u />x^{2} + 12 x = 48<u />

in the form ax^{2}  +bx +c = 0

= x^{2} +12x - 48

make use of the formula :

\frac{-b+-\sqrt{b^{2} -4ac} }{2a}

replace values to make 2 equations :

1.\frac{-12+\sqrt{12^{2} -4*1*-48} }{2*1} = 3.17

2.\frac{-12-\sqrt{12^{2} -4*1*-48} }{2*1} = -15.2

hence <u>x = 3.17 and x = -15.2</u>

<u />

<h2><u>Question 3 </u></h2>

<u />x^{2} -14x+40=0<u />

use product and sum method

product = 40

sum = -14

numbers needed = (-10 , -4)

x - 10 = 0

x - 4 = 0

hence<u> x = 10 and x = 4</u>

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<h2><u>Question 4 </u></h2>

<u />5b^{2} -20b-18 = 7<u />

in the form ax^{2}  +bx +c = 0

this becomes 5b^{2} -20b-18-7

= 5b^{2} -20b-25

can simplify by 5

= b^{2} -4b-5 =0\\

use product and sum method

product = -5

sum = -4

numbers needed (-5 , 1)

b-5 = 0

b + 1 = 0

hence <u>b = 5 and b = -1</u>

5 0
3 years ago
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