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3 years ago

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]

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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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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 (h, k) is the center and r is the radius.

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

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

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

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

Solve for r:

$(-5)^2+(3)^2=r^2$

Simplify. Thus:

$25+9=r^2$

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:

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

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3 years ago