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3 years ago

Fill in the table using this function rule. y=28 - 3x X Y 1 3 4 6

Mathematics

1 answer:

My name is Ann [436]3 years ago

4 0

Answer:

So, you just end up plugging in your numbers.

y = 28 - 3(1)

y = 28-3

y = 25

Y = 28 - 3(3)

y = 28 - 9

y = 19

y = 28 - 3x

y = 28 - 3(4)

y = 28 - 12

y = 16

y = 28 - 3x

y = 28 - 3(6)

y = 28 - 18

y = 10

Step-by-step explanation:

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I need to know if this H(3,2),J(4,1),K(-2,-4),M(-1,-5) is parallel perpendicular or neither I need to show my work and write my

Shkiper50 [21]

Answer:

Step-by-step explanation:

H(3,2) & J(4, 1)

$Slope = \dfrac{y_{2} -y_{1}}{x_{2}-x_{1}}\\\\=\dfrac{1-2}{4-3}\\\\\\=\dfrac{-1}{1}\\\\= -1$

K(-2,-4) & M(-1 , -5)

$Slope =\dfrac{-5-[-4]}{-1-[-2]}\\\\=\dfrac{-5+4}{-1+2}\\\\= \dfrac{-1}{1}\\\\\\= -1$

Line HJ and KM have same slopes. So, they are parallel

6 0

2 years ago

The sum of two numbers is 48.<br>the second number is seven times the first number

dusya [7]

Answer:

$\huge\boxed{\sf x =6,\ y = 42}$

Step-by-step explanation:

Let the numbers be x and y

<h3>Given condition:</h3>

x + y = 48 --------(1)

y = 7x -------------(2)

Put Eq. (2) in (1)

x + 7x = 48

8x = 48

Divide 8 to both sides

x = 48/8

Put x = 6 in Eq. (2)

y = 7 (6)

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6 0

1 year ago

Read 2 more answers

30 POINTS AVAILABLE

kherson [118]

Answer:

$\large\boxed{(x-2)^2+(y-1)^2=34}$

Step-by-step explanation:

The equation of a circle in standard form:

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

(h, k) - center

r - radius

We have the endpoints of the diameter: (-1, 6) and (5, -4).

Midpoint of diameter is a center of a circle.

The formula of a midpoint:

$\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)$

Substitute:

$h=\dfrac{-1+5}{2}=\dfrac{4}{2}=2\\\\k=\dfrac{6+(-4)}{2}=\dfrac{2}{2}=1$

The center is in (2, 1).

The radius length is equal to the distance between the center of the circle and the endpoint of the diameter.

The formula of a distance between two points:

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

Substitute the coordinates of the points (2, 1) and (5, -4):

$r=\sqrt{(5-2)^2+(-4-1)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}$

Finally we have:

$(x-2)^2+(y-1)^2=(\sqrt{34})^2$

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3 years ago

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Helga [31]

Answer: 17

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3 years ago

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