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Snezhnost [94]
3 years ago
8

The quadratic x? - x = 2 has two solutions. What is the larger of the two solutions?

Mathematics
1 answer:
chubhunter [2.5K]3 years ago
5 0
I believe the 2 is the largest
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Find the vertex of the function given below. y = x2 - 4x + 1
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The vertex is (2, -3).
Use the formula x= -b/2a to find the x value., which is 2.Then, substitute it into the function, wherever you see an x. Follow order operations and you get y, which is -3.
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Find the area of a circle with a diameter of 16 inches. Use 3.14 for pi.
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200.96‬

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A = πr²

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The graph shows how a motorboat travels around a lake. What does the graph most likely show?
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3 years ago
Question is on picture attached
nalin [4]

7(\frac{9}{4}x- \frac{18}{7})= -\frac{5}{21} +\frac{17}{7}x   \frac{63}{4}x-18 =-\frac{5}{21}+ \frac{17}{7}x\\ \frac{63}{4}x -  \frac{17}{7}x = 18-\frac{5}{21}\frac{373}{28} =\frac{373}{21} x=\frac{4}{3}

ok done. Thank to me :>

5 0
2 years ago
Using the binomial theorem , obtain the expansion of :
andrezito [222]

Answer:

see explanation

Step-by-step explanation:

Expand both factors and collect like term

Using Pascal' triangle with n = 6 to obtain the coefficients

1  6  15  20  15  6  1

Decreasing powers of 1 from 1^{6} to 1^{0}

Increasing powers of 3x from (3x)^{0} to (3x)^{6}

1+3x)^{6}

= 1.1^{6}(3x)^{0} + 6.1^{5}(3x)^{1} + 15.1^{4}(3x)^{2} + 20.1^{3}(3x)^{3} + 15.1²(3x)^{4} + 6.1^{1}(3x)^{5} + 1.1^{0}(3x)^{6}

= 1 + 18x + 135x² + 540x³ + 1215x^{4} + 1458x^{5} + 729x^{6}

--------------------------------------------------------------------------------------

(1-3x)^{6}

= 1.1^{6}(-3x)^{0} + 6.1^{5}(-3x)^{1} + 15.1^{4}(-3x)^{2} + 20.1^{3}(-3x)^{3} + 15.1²(-3x)^{4} + 6.1^{1}(-3x)^{5} + 1.1^{0}(-3x)^{6}

= 1 - 18x + 135x² - 540x³ + 1215x^{4} - 1458x^{5} + 729x^{6}

----------------------------------------------------------------------------------

Collecting like terms from both expressions

(1+3x)^{6} + (1-3x)^{6}

= 2 + 270x² + 2430x^{4} + 1458x^{6}

----------------------------------------------------

(2)

Using Pascal's triangle with n = 5

1  5  10  10  5  1

Decreasing powers of 1 from 1^{5} to 1^{0}

Increasing powers of 2x from (2x)^{0} to (2x)^{5}

(1+2x)^{5}

= 1.1^{5}(2x)^{0} + 5.1^{4}(2x)^{1} + 10.1^{3}(2x)^{2} + 10.1^{2}(2x)^{3} + 5.1^{1}(2x)^{4}+ 1.1^{0}(2x)^{5}

= 1 + 10x + 40x² + 80x³ + 80x^{4} + 32x^{5}

8 0
3 years ago
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