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Delvig [45]
3 years ago
7

Suppose a parabola has an axis of symmetry at x = -5, a maximum height of 9, and passes through the point (-7,1). write an equat

ion of the parabola in vertex from.
Y = -4(x-5)^2+9
Y = -7(x+9)^2 - 5
Y = -0.06(x-5)^2 + 9
Y = -2(x+5)^2 + 9

Help please...
Mathematics
1 answer:
frosja888 [35]3 years ago
4 0

We are given

a parabola has an axis of symmetry at x = -5, a maximum height of 9

so, we get

vertex =(-5,9)

vertex=(h,k)=(-5,9)

so, h=-5 and k=9

we can use vertex form of parabola

y=a(x-h)^2+k

we can plug these value

y=a(x+5)^2+9

now, it passes through the point (-7,1)

we can use it and then we can solve for a

1=a(-7+5)^2+9

4a+9=1

4a=-8

a=-2

So, we will get equation of parabola as

y=-2(x+5)^2+9..............Answer


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Option B: Approximately bell-shaped

Explanation:

Option A: Skewed

A histogram is said to be skewed if most of the larger values falls either on the left or right side.

Clearly, the given histogram is not skewed.

Thus, Option A is not the correct answer.

Option B: Approximately bell-shaped

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Option C: Uniform

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Option D: Skewed left

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5 0
3 years ago
Read 2 more answers
When you put your little brother in the dryer and say spinjitzu is the dryer supposed to leak blood
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No

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6 0
3 years ago
NO SCAM PLEASEEEE
damaskus [11]
The answer is 11/36

2/12 chance of rolling fours

because there are 2 sides containing a four on both dice combined and 12 sides in total.

Doubles mean you have to roll the same number simultaneously so let’s say we want to calculate the probability for double ones: then it’s 1/6 on the first dice for a one, and 1/6 on the second dice to land on a one as well.

I personally like to imagine a box like this:
_ _ _ _ _ _
|
|
|
|
|
|

If you have one dice then it’s just a random segment on one of the lines. If you want the specific result from two dice then you want two specific segments which is also the 1 specific tile out of 36 (6 width times 6 height). So you multiply.

1/6 * 1/6 = 1/36 chance to roll double of ones

And 1/36 chance to roll double twos, threes, fours, fives, and sixes. But we don’t count the double fours because any four will do. So:

1/36 * 5 = 5/36

So for the probability of either doubles or containing a four is the probability of doubles of either number plus the probability of either dice being a four:

5/36 + 2/12 =

5/36 + 6/36 =

11/36
5 0
2 years ago
The 12th term and the sim of 1+3+5+7+.....
zzz [600]

To find this, first find the factor or rate of which the numbers are moving. To do so do as follows.

subtract 1 from 3

3-1=2

So each number is having 2 added to it.

Now add two to 7 and the numbers afterwards till you get the 12th term

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1+3+5+7+9

9+2=11

1+3+5+7+9+11

11+2=13

1+3+5+7+9+11+13

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19+2=21

1+3+5+7+9+11+13+15+17+19+21

21+2=23

1+3+5+7+9+11+13+15+17+19+21+23

So 23 is the 12th term


7 0
3 years ago
What is 321×23 <br><br> one more thing<br><br> what is 3,829÷1221
Archy [21]
321 times 23 is 7,383
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(the real number was: 3.13595414)
3 0
3 years ago
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