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aksik [14]
3 years ago
12

Re-write the quadratic function below into standard form y=-4(x-1)(x-1)-3

Mathematics
1 answer:
vodka [1.7K]3 years ago
3 0

Answer:

y= -4x2+8x-7

Step-by-step explanation:

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HACTEHA [7]

Answer:

the anwer is B ( i mean second option)

And you can try it

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y =  \frac{x}{3}  - 1

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Step-by-step explanation:

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5 0
3 years ago
You are given a number line with a left endpoint R, a point in between the two endpoints (not a midpoint though) of S, and a rig
Nina [5.8K]
You just add the lengths of RS and ST together. In your case RT= 24
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3 years ago
Which point is not part of circle C?<br><br> C<br> P<br> Q<br> R
Juliette [100K]

bearing in mind that a <u>circle</u> means the round thing, so only the points on the round thing are ON the circle, others like the RC chord or the PQ diameter or the point C are NOT ON the circle.

4 0
4 years ago
Read 2 more answers
Victoria spent one-fourth of her birthday money on clothes. She received another $25 a week later. If she has a total of $70 now
tia_tia [17]

Answer:

she had $60 before she went for shopping

Step-by-step explanation:

PLZ MARK BRAINLIEST

Let x represent the amount of money that Victoria had before she went for shopping.

Victoria spent one-fourth or her birthday money on clothes. It means that the amount she spent on shopping is 1/4 × x = x/4. Amount that she was having left would be x - x/4 = 3x/4

She received another 25$ a week later. The amount that she is having at this point will be 3x/4 + 25

If she has a total of 70$ now, it means that

3x/4 + 25 = 70

Multiplying through by 4

3x + 100 = 280

3x ,= 280 - 100 = 180

x = 180/3 = 60

3 0
3 years ago
What is the probability of drawing a 3 or a 4 or a heart from a deck of cards?
Effectus [21]
\Omega=\{2\spadesuit;\ 3\spadesuit;...;A\spadesuit;2\heartsuit ;\ 3\heartsuit;...;A\heartsuit;\ 2\diamondsuit;\ 3\diamondsuit;...;A\diamondsuit;\ 2\clubsuit;\ 3\clubsuit;...;A\clubsuit\}\\\\|\Omega|=52\\\\A=\{3\spadesuit;\ 4\spadesuit;\ 3\diamondsuit;\ 4\diamondsuit;\ 3\clubsuit;\ 4\clubsuit;\ 2\heartsuit;\ 3\heartsuit;\ 4\heartsuit;...;A\heartsuit\}\\\\|A|=2+2+2+13=19\\\\P(A)=\dfrac{|A|}{|\Omega|}\to P(A)=\dfrac{19}{52}

Answer:\ \dfrac{19}{52}
5 0
3 years ago
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