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Radda [10]
3 years ago
14

What transformation occurs to f(x) in the function f(x)-7

Mathematics
1 answer:
pshichka [43]3 years ago
5 0

Answer:

Left

Step-by-step explanation:

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Show that line passing through points (3,-2) and (-4,-2) is a horizontal line​
Ierofanga [76]

Step-by-step explanation:

horizontal lines will have a gradient(m) of 0

so if m=0 , the line is horizontal

formula to find m:

m=y2-y1 / x2-x1

insert coordinates into formula

m =  \frac{ - 2 - ( - 2)}{ - 4 - 3}  \\  \\ m =  \frac{ - 2 + 2}{ - 7}  \\ m =  \frac{0}{ - 7}  \\ m = 0

As m=0 , therefore it is a horizonta line.

hope this helps..

3 0
2 years ago
Which of the following is equal to 7 1/3?
earnstyle [38]

Answer:

C

Step-by-step explanation:

If the number's a fraction, which this one is, the denominator will always be that little number by the radical, and the whole number will be in the root. Hope this helps!

5 0
3 years ago
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The constant of proportionality of chicken nuggets in relation
zhuklara [117]

I beleive that it would be C=n25

Step-by-step explanation:

I'm not sure if this is not complex enough but I'm just going by what I think this is.

4 0
3 years ago
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Solve the following system of equations.
zheka24 [161]

Answer:

(1,1)

Step-by-step explanation:

8 0
3 years ago
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Which equation represents a circle that contains the point (-5, 3) and has a center at (-2, 1)? Distance formula: vaa -02 (x - 1
natali 33 [55]

Given:

The center of the circle = (-2,1).

Circle passes through the point (-5,3).

To find:

The equation of the circle.

Solution:

Radius is the distance between the center of the circle and any point on the circle. So, radius of the circle is the distance between the points (-2,1) and (-5,3).

Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

r=\sqrt{(-5-(-2))^2+(3-1)^2}

r=\sqrt{(-5+2)^2+(2)^2}

r=\sqrt{(-3)^2+(2)^2}

On further simplification, we get

r=\sqrt{9+4}

r=\sqrt{13}

The standard form of a circle is:

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

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

Substitute h=-2, k=1 and r=\sqrt{13}.

(x-(-2))^2+(y-1)^2=(\sqrt{13})^2

(x+2)^2+(y-1)^2=13

Therefore, the equation of the circle is (x+2)^2+(y-1)^2=13.

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