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sergey [27]
3 years ago
8

Question 1 of 10

Mathematics
1 answer:
Archy [21]3 years ago
3 0

Answer:

A. y = 7(x + 1)²-3

Step-by-step explanation:

Parabola:

y = 7x^{2} + 14x + 4

y = 7(x^{2} + 2x) + 4

Putting into vertex form, remember that:

(x + a)^{2} = x^{2} + 2ax + a^{2}

In this question:

x^{2} + 2x, to put into this format:

x^{2} + 2x + 1 = (x + 1)^{2}

We add one inside the parenthesis to do this. The parenthesis is multiplied by 7, so for the equivalent, we also have to subtract 7. Then

Vertex form:

y = 7(x^{2} + 2x + 1) + 4 - 7

y = 7(x + 1)^{2} - 3

So the correct answer is:

A. y = 7(x + 1)²-3

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Answer:

32

Step-by-step explanation:

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Find the area of the regular pentagon. 4.1 cm 6 cm Area = [?] cm? Enter your answer to the nearest tenth. ​
Assoli18 [71]

Answer:

Area of the given regular pentagon is 61.5 cm².

Step-by-step explanation:

Area of a regular polygon is given by,

Area = \frac{1}{2}aP

Here, a = Apothem of the polygon

P = Perimeter of the polygon

Apothem of the regular pentagon given as 4.1 cm.

Side of the pentagon = 6 cm

Perimeter of the pentagon = 5(6)

                                             = 30 cm

Substituting these values in the formula,

Area = \frac{1}{2}(4.1)(30)

        = 61.5 cm²

Therefore, area of the given regular pentagon is 61.5 cm².

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3 years ago
The ratio table below shows the relationship between the weight of apples purchased and the total cost of the apples. Apple Cost
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3 years ago
Three stamps can be attached to each other in various ways.how many ways might three stamps be attached?
ivanzaharov [21]
Let's call the stamps A, B, and C. They can each be used only once. I assume all 3 must be used in each possible arrangement.
There are two ways to solve this. We can list each possible arrangement of stamps, or we can plug in the numbers to a formula. 
Let's find all possible arrangements first. We can easily start spouting out possible arrangements of the 3 stamps, but to make sure we find them all, let's go in alphabetical order. First, let's look at the arrangements that start with A:
ABC
ACB
There are no other ways to arrange 3 stamps with the first stamp being A. Let's look at the ways to arrange them starting with B:
BAC
BCA
Try finding the arrangements that start with C:
C_ _
C_ _
Or we can try a little formula; y×(y-1)×(y-2)×(y-3)...until the (y-x) = 1 where y=the number of items.
In this case there are 3 stamps, so y=3, and the formula looks like this: 3×(3-1)×(3-2).
Confused? Let me explain why it works.
There are 3 possibilities for the first stamp: A, B, or C. 
There are 2 possibilities for the second space: The two stamps that are not in the first space.
There is 1 possibility for the third space: the stamp not used in the first or second space. 
So the number of possibilities, in this case, is 3×2×1.
We can see that the number of ways that 3 stamps can be attached is the same regardless of method used.


8 0
3 years ago
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