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Sati [7]
3 years ago
9

Ayden has $0.56 worth of pennies and nickels. He has a total of 20 penni

Mathematics
1 answer:
elena55 [62]3 years ago
8 0

Answer:

9 nickels and 11 pennies

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H(x) = 3x-4 what is h(6)
lina2011 [118]

Answer:

14

Step-by-step explanation:

3*6=18

18-4=14

8 0
3 years ago
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Use the x-intercept method to find all real solutions of the equation.<br> x^3-10x^2+29x-20=0
stiks02 [169]

Answer:

X=1,5,4

Step-by-step explanation:

3 0
3 years ago
HELP PLEASE! I would really appreciate it :)
dem82 [27]
Each time x increases by 2 (eg: from x = -3 to x = -1), the value of y increases by 7 (eg: y = 9.5 to y = 16.5)

Change in y = 7
change in x = 2
Slope = (change in y)/(change in x) = 7/2 = 3.5

This all means that we have a linear equation
4 0
3 years ago
The area of a rectangle is 56 cm. The length is 2 cm more than x and the width is 5 cm less than twice x. Solve for x. Round to
Black_prince [1.1K]
 Length = x + 2  (because it is 2 cm more than x)
Width = 2x - 5   (5 cm lest than 2x)Area = 54 cm2  this is the formula to find the area Length × Width = Area (x + 2)(2x - 5) = 542x2 - x - 10 = 54   (this is you area)

  Subtract 54 on both sides of equation to make the right side zero. 2x2 - x - 64 = 0  then use the quadratic formula x = (-b ± √(b2 - 4ac)) / 2a where:a = 2b = -1c = -64 Plug in these values into the formula. x = (1 ± √(1 - 4(-128))) / 4 x = (1 ± √(513)) / 4 x = (1 ± 22.65) / 4 x = (1 + 22.65) / 4          and          x = (1 - 22.65) / 4 x = 5.91                          and          x = -5.41  Check the validity of the x values by adding them to the length and width.  If the length or width should be a negative value, then that value of x is not acceptable. Now x = 5.91 Length = 5.91 + 2  (positive value.)Width = 2(5.91) - 5  ( positive value.) x = 5.91    If we look at this -- x = -5.41, Both length and width will be negative values. We reject this value of x.  The answer is x = 5.91

Hope I helped and sorry it was really long
6 0
2 years ago
Read 2 more answers
Which expressions are equivalent to the one below. Check all that apply. 16^x
madam [21]

A.\ 4^x\cdot4^x=4^{x+x}=4^{2x}\qquad\text{used}\ a^n\cdot a^m=a^{n+m}\\\\B.\ 4^{2x}=(4^2)^x=16^x\qquad\text{used}\ (a^n)^m=a^{nm}\\\\C.\ 4^2\cdot4^x=4^{2+x}\\\\D.\ (4\cdot4)^x=16^x\\\\E.\ 4\cdot4^{2x}=4^{1+2x}\\\\F.\ 4\cdot4^x=4^{1+x}\\\\Answer:\ \boxed{B.\ and\ D.}

3 0
3 years ago
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