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lana [24]
3 years ago
13

Which pair of equations generates graphs with the same vertex?

Mathematics
2 answers:
Mama L [17]3 years ago
7 0
A quadratic equation may be expressed in vertex form as:
y = a(x - h)² + k, the vertex being (-h , k)
For option A, the vertices are visibly different. (-4 , 0) and (4 , 0)
For B, the graphs' vertices are the same, (0 , 0).
For C, the vertices (can be obtained by differentiating and putting equal to 0) are (0 , -4) and (0 , 4)
For D, the vertices are (4 , 0) and (0 , 4)
Thus, the answer is B.
lesantik [10]3 years ago
5 0

Answer:

B)

Step-by-step explanation:

Edg 2020

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The standard form of the equation that represents the number of quarters, q, and the number of dimes, d, that Austin has in his
vazorg [7]

The standard form of the equation that represents the number of quarters, q, and the number of dimes, d, that Austin has in his piggy bank is 5q + 2d = 270. This can be change to slope intercept form to see the clearer picture. By dividing the whole equation by 2, so the equation become d= -2.5q + 135.

<span>This means that Austin has initially 135 dimes in his piggy bank and he is losing 2.5 dimes per quater</span>

8 0
3 years ago
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yan [13]

Answer:

D, Impossible

Step-by-step explanation:

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3 years ago
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F(x) = x^2. What is g(x)?
Ivan
B. hope I helped have a good day !
3 0
3 years ago
The numbers 1, 2, 3, 4,5, 6, 7, 8, 9 are arranged in a list so that each number is either greater than all the numbers that come
Nina [5.8K]

The lists of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 that are possible is; 256

<h3>How to Solve Probability Combinations?</h3>

We are told that the last digit has to be either a one (1) or nine (9), then are two (2) options (>,<) for the remaining eight digits, so there are 2⁸ = 256 arraignments where each digit is either greater than or less than the preceding digits. This can be confirmed as below;

Numbers beginning with 1 = 8C0 = 1

Numbers beginning with 2 = 8C1 = 8

Numbers beginning with 3 = 8C2 = 28

Numbers beginning with 4 = 8C3 = 56

Numbers beginning with 5 = 8C4 = 70

Numbers beginning with 6 = 8C5 = 56

Numbers beginning with 7 = 8C6 = 28

Numbers beginning with 8 = 8C7 = 8

Numbers beginning with 9 = 8C8 = 1

Total lists of numbers possible = 1 + 8 + 28 + 56 + 70 + 56 + 28 + 8 + 1 = 256

Read more about Probability Combinations at; brainly.com/question/25688842

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4 0
2 years ago
HELP QUICK 10 POINTS!!!!!!
padilas [110]
\frac{x}{5} = 10   Multiply both sides by 5 to cancel out the fraction
x = 50
5 0
3 years ago
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