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Juli2301 [7.4K]
3 years ago
7

X/4 + 9 = x ? A)12, B)14, C)10, D)5​

Mathematics
2 answers:
olga_2 [115]3 years ago
8 0

Answer:

A)12 is the answer.....

Eddi Din [679]3 years ago
5 0

Answer:

A) 12.

Step-by-step explanation:

Plug in 12 for x in the given equation:

x/4 + 9 = x

(12)/4 + 9 = 12

Simplify. Remember to follow PEMDAS. First, divide, then add:

(12/4) + 9 = 12

(3) + 9 = 12

12 = 12 (True).

~

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\text{Hello there! :)}

Answer:

\boxed{\text{Center at } (7, -4), \text{ Radius of } 7.}

\text{Formula for a circle: }  (x - h)^{2} + (y - k)^{2}  = r^{2}  \text{ Where:}

h = \text{ x-coordinate of center}\\\\k = \text{ y-coordinate of center}\\\\r = \text{ radius}\\\\\text{Therefore:}

\text{In the equation } (x - 7)^{2}  + (y + 4)^{2}  = 49:

h = 7\\\\k = -4\\\\r = \sqrt[]{49} = 7 \\\\\text{So:}

\text{Center at } (7, -4), \text{ Radius of } 7.

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3 years ago
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* There are five cars in line at a stoplight. Each of these cars is a different color and different type of car. Diego is drivin
Svetach [21]
Info given :
vehicles : (colors)red car, green car, yellow car, blue car, white car
types : sedan, minivan, SUV, convertible, pick-up truck
people : Diego, Ingrid, Rachel, Anton, Yuna

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Ingrid -- white car -- sedan -- 4th car
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Rachel -- yellow car -- SUV -- 1st car
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I am thinking the truck, it is green, is driven by Yuna
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A doll factory needed to produce an order of dolls for a shop. The factory
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24 / 1.6 = 15

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How do I solve this
MatroZZZ [7]

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36

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A circle is centered at J(3, 3) and has a radius of 12.
stealth61 [152]

Answer:

(-6,\, -5) is outside the circle of radius of 12 centered at (3,\, 3).

Step-by-step explanation:

Let J and r denote the center and the radius of this circle, respectively. Let F be a point in the plane.

Let d(J,\, F) denote the Euclidean distance between point J and point F.

In other words, if J is at (x_j,\, y_j) while F is at (x_f,\, y_f), then \displaystyle d(J,\, F) = \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}.

Point F would be inside this circle if d(J,\, F) < r. (In other words, the distance between F\! and the center of this circle is smaller than the radius of this circle.)

Point F would be on this circle if d(J,\, F) = r. (In other words, the distance between F\! and the center of this circle is exactly equal to the radius of this circle.)

Point F would be outside this circle if d(J,\, F) > r. (In other words, the distance between F\! and the center of this circle exceeds the radius of this circle.)

Calculate the actual distance between J and F:

\begin{aligned}d(J,\, F) &= \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}\\ &= \sqrt{(3 - (-6))^{2} + (3 - (-5))^{2}} \\ &= \sqrt{145}  \end{aligned}.

On the other hand, notice that the radius of this circle, r = 12 = \sqrt{144}, is smaller than d(J,\, F). Therefore, point F would be outside this circle.

5 0
3 years ago
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