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kolbaska11 [484]

3 years ago

12

Last year, Kelly's Candies sold 425,524,812 candy bars. Chester's Chocolates sold 425,520,736 candy bars. Which company sold mor

e candy bars?

2 answers:

sweet-ann [11.9K]3 years ago

6 0

Kelly's Candies because 425,524,812 is the bigger number.

stepladder [879]3 years ago

3 0

Kelly's Candies because 425,524,812 is bigger that 425,520,736

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Match the parabolas represented by the equations with their vertices. y = x2 + 6x + 8 y = 2x2 + 16x + 28 y = -x2 + 5x + 14 y = -

GaryK [48]

Consider all parabolas:

1.

$y = x^2 + 6x + 8,\\y=x^2+6x+9-9+8,\\y=(x^2+6x+9)-1,\\y=(x+3)^2-1.$

When x=-3, y=-1, then the point (-3,-1) is vertex of this first parabola.

2.

$y = 2x^2 + 16x + 28=2(x^2+8x+14),\\y=2(x^2+8x+16-16+14),\\y=2((x^2+8x+16)-16+14),\\y=2((x+4)^2-2)=2(x+4)^2-4.$

When x=-4, y=-4, then the point (-4,-4) is vertex of this second parabola.

3.

$y =-x^2 + 5x + 14=-(x^2-5x-14),\\y=-(x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}-14),\\y=-((x^2-5x+\dfrac{25}{4})-\dfrac{25}{4}-14),\\y=-((x-\dfrac{5}{2})^2-\dfrac{81}{4})=-(x-\dfrac{5}{2})^2+\dfrac{81}{4}.$

When x=2.5, y=20.25, then the point (2.5,20.25) is vertex of this third parabola.

4.

$y =-x^2 + 7x + 7=-(x^2-7x-7),\\y=-(x^2-7x+\dfrac{49}{4}-\dfrac{49}{4}-7),\\y=-((x^2-7x+\dfrac{49}{4})-\dfrac{49}{4}-7),\\y=-((x-\dfrac{7}{2})^2-\dfrac{77}{4})=-(x-\dfrac{7}{2})^2+\dfrac{77}{4}.$

When x=3.5, y=19.25, then the point (3.5,19.25) is vertex of this fourth parabola.

5.

$y =2x^2 + 7x +5=2(x^2+\dfrac{7}{2}x+\dfrac{5}{2}),\\y=2(x^2+\dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x^2+\dfrac{7}{2}x+\dfrac{49}{16})-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x+\dfrac{7}{4})^2-\dfrac{9}{16})=2(x+\dfrac{7}{4})^2-\dfrac{9}{8}.$

When x=-1.75, y=-1.125, then the point (-1.75,-1.125) is vertex of this fifth parabola.

6.

$y =-2x^2 + 8x +5=-2(x^2-4x-\dfrac{5}{2}),\\y=-2(x^2-4x+4-4-\dfrac{5}{2}),\\y=-2((x^2-4x+4)-4-\dfrac{5}{2}),\\y=-2((x-2)^2-\dfrac{13}{2})=-2(x-2)^2+13.$

When x=2, y=13, then the point (2,13) is vertex of this sixth parabola.

3 0

2 years ago

How do I solve this question?

Sati [7]

Answer:

m=2 and n=3

Step-by-step explanation:

<u>Step</u> :-

Given $[ 2 x^{n}y^{2} ]^m = 4 x^6 y^4$

using algebraic formula $(a^m)^n= a^{mn}$

now

$2^{m} x^{mn} y^{2m} = 4x^{6} y^{4}$

now equating 'x' powers, we get

$x^{mn} = x^{6}$

$mn=6$ ....(1)

now

$y^{2m} = y^{4}$

Equating 'y' powers ,we get

2 m=4

m=2

substitute m= 2 in equation (1)

we get

2 n=6

n=3

verification:-

substitute m=2 and n=3 , we get

$[ 2 x^{n}y^{2} ]^m = 4 x^6 y^4$

$(2 x^{3} y^2)^2 = 4 x^6 y ^4\\$

$4 x^6 y^4 = 4 x^6 y^4$

both are equating so m= 2 and n=3

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2 years ago

List the factors in decimals in order from least to greatest 0.5 3/16 0.75. 5/48

suter [353]

5/48
3/16
0.5
0.75

This is listen in least to greatest...

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2 years ago

Please help ASAP :ppppp

Morgarella [4.7K]

Answer:

I think it's c

Step-by-step explanation:

6 0

2 years ago

Solve (10b+7b^2-6b^3)-(12b^2-3b^3)

MrRa [10]

Answer:

-3b³ - 5b² + 10b

General Formulas and Concepts:

<u>Pre-Algebra</u>

Distributive Property

<u>Algebra I</u>

Combining Like Terms

Step-by-step explanation:

<u>Step 1: Define Expression</u>

(10b + 7b² - 6b³) - (12b² - 3b³)

<u>Step 2: Simplify</u>

Distribute negative: 10b + 7b² - 6b³ - 12b² + 3b³
Combine like terms (b³): -3b³ + 10b + 7b² - 12b²
Combine like terms (b²): -3b³ - 5b² + 10b

3 0

2 years ago