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Romashka-Z-Leto [24]
3 years ago
8

If the diameter of a circle has endpoints A (7,2) and B (-1,8), where is the center?

Mathematics
2 answers:
Elenna [48]3 years ago
6 0
<span>first we will get sum of X coordinates of A and B
secondly we will divide it by two
same sum the Y coordinates and divide it by
(-1+7)/2=3 and  (2+8)/2=5
so the center will be (3,5)
hope it helps
</span>
docker41 [41]3 years ago
3 0
In the middle: x-coordinate= {7-(-1)}/2= 4 y-coordinate= (8-2)/2= 3 So centre is (4,3)
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Find the slope of the line containing the two points.<br><br> (2,5),(-3,-2)<br><br> Please show work
HACTEHA [7]

Answer:

7/5

Step-by-step explanation:

To solve this problem we will use the formula to find the slope of a line given two points (x₁, y₁), (x₂, y₂)

The formula is:

m= \frac{y{_2} -y_{1}}{x_{2}-x_{1}  } \\\\

So, we're going to use the points (2,5), (-3, -2) in this formula.

m= \frac{-2-(5)}{-3-(2)} \\m=\frac{-2-5}{-3-2}\\m=\frac{-7}{-5} \\m=\frac{7}{5}

Therefore, the slope is 7/5

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3 years ago
7. Virus growth is an example of which type of function?
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3 years ago
Given the following diagram, find the missing measure.<br>*​
BigorU [14]

Option C:

x = 30

Solution:

The given image is a triangle.

angle 1, angle 2 and angle 3 are interior angles of a triangle.

angle 4 is the exterior angle of a triangle.

m∠4 = 2x°, m\angle2=\frac{4}{3}x^\circ, m∠3 = 20°

Exterior angle theorem:

<em>In triangle, the measure of exterior angle is equal to the sum of the opposite interior angles.</em>

By this theorem,

m∠4 = m∠2 + m∠3

$2x^\circ=\frac{4}{3}x^\circ+20^\circ

Subtract \frac{4}{3}x^\circ on both sides of the equation.

$2x^\circ-\frac{4}{3}x^\circ=20^\circ

To make the denominator same and then subtract.

$\frac{6}{3}x^\circ -\frac{4}{3}x^\circ=20^\circ

$\frac{2}{3}x^\circ =20^\circ

Multiply by \frac{3}{2} on both sides of the equation.

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6 0
3 years ago
Please answer this correctly
Dahasolnce [82]

Answer:

9 5/7

Step-by-step explanation:

You add all the 13/14 then divided it by 14

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8 0
3 years ago
En una fiesta hay 40 asistentes de los cuales el número de varones que no bailan es el doble que el número de las mujeres que no
Arte-miy333 [17]

Answer:

50%

Step-by-step explanation:

Para resolver este problema vamos a definir 4 variables:

V= número de varones bailando.

M= número de mujeres bailando.

V' = número de varones que no están bailando.

M' = número de mujeres que no están bailando.

Sabemos que hay 40 personas en la fiesta, por lo que podemos construir la siguiente ecuación:

V+M+V'+M'=40

Vamos a suponer que cada mujer que está bailando, está bailando varón respectivamente. (El problema no da más información, entonces podemos suponer esto.)

Entonces si hay 5 mujeres bailando, entonces también hay 5 hombres bailando, por lo que nuestra ecuación se reescribe de la siguiente manera:

5+5+V'+M'=40

y simplificamos:

10+V'+M'=40

V'+M'=40-10

V'+M'=30

Ahora bien, el problem nos dice que el número de varones que no bailan es el doble del número de mujeres que no bailan en un determinado momento. Entonces con esta información podemos construir la siguiente ecuación:

V'=2M'

Y podemos despejar el número de mujeres que no bailan, lo que nos da:

M'=\frac{V'}{2}

Entonces podemos sustituir esto dentro de nuestra ecuación para obtener:

\frac{V'}{2}+V'=30

y podemos entonces despejar V'

\frac{3V'}{2}=30

V'=\frac{2(30)}{3}

V'=20

Entonces hay 20 varones que no están bailando, por lo que la probabilidad de que el varón que se escoge al azar no esté bailando está dada por la siguiente fórmula:

P=\frac{V'}{total}

P=\frac{20}{40}=\frac{1}{2}

P=50%

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