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Kaylis [27]
3 years ago
7

Given that

Mathematics
2 answers:
GarryVolchara [31]3 years ago
6 0

f(x) = x^2

1st transpose: f(x) = x^2 +7

2nd dilate: f(x) = 9(x^2+7)

3rd reflect: f(x) = 9x^2 + 63


zheka24 [161]3 years ago
4 0

f(x) = x^2

shift f(x) up 7 units means adding 7 to f(x)

so now f(x) = x^2 +7

vertically stretch the function by a factor of 9 means multiplying f(x) by 9

so now f(x) = 9(x^2+7)

f(x) = 9x^2 + 63

reflect across the y-axis means x becomes -x in f(x)

so now f(x) = 9(-x)^2 + 63

f(x) = 9x^2 + 63


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Mrac [35]

Answer:

y=2(x+10)

Step-by-step explanation:

its a simple function :)

if y is the weight of the bag, and there is already a base of 10, the function explained would be

2* x <-- (the amount of days) +10 <- the amount already in there

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3 years ago
Is 1.2 less than or higher than 1.5
Olenka [21]

1.2 is less than 1.5

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4 0
3 years ago
Find the equation of a circle with a center at (7,2) and a point on the circle at (2,5)?
monitta

Answer:

(x-7)^2+(y-2)^2=34

Step-by-step explanation:

We want to find the equation of a circle with a center at (7, 2) and a point on the circle at (2, 5).

First, recall that the equation of a circle is given by:

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

Where (<em>h, k</em>) is the center and <em>r</em> is the radius.

Since our center is at (7, 2), <em>h</em> = 7 and <em>k</em> = 2. Substitute:

(x-7)^2+(y-2)^2=r^2

Next, the since a point on the circle is (2, 5), <em>y</em> = 5 when <em>x</em> = 2. Substitute:

(2-7)^2+(5-2)^2=r^2

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<em />(-5)^2+(3)^2=r^2<em />

Simplify. Thus:

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Finally, add:

r^2=34

We don't need to take the square root of both sides, as we will have the square it again anyways.

Therefore, our equation is:

(x-7)^2+(y-2)^2=34

4 0
3 years ago
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