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3 years ago

Y = 6x - 4; x = 8 find the value of y for the given value of x

Mathematics

2 answers:

Greeley [361]3 years ago

7 0

You put the value of x in the given equation to get the value of Y.
Y = 6(8) - 4
Y = 44

miskamm [114]3 years ago

5 0

X=8
y=6x-4
y=6(8)-4
y=48-4
y=44

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SOLVE 6-9. State of each triangle is a right triangle

11111nata11111 [884]

To find out if a triangle is a right triangle, you can use the Pythagorean theorem(which can only be used for right triangles):

a² + b² = c² (c is the hypotenuse or the longest side) And you can plug in the side lengths into this equation. If they are the same number on both sides, it is a right triangle, if they are different numbers it is not a right triangle.

6.) a² + b² = c²

(4√3)² + (11)² = (13)²

(16(3)) + 121 = 169

48 + 121 = 169

169 = 169 It IS a right triangle

7.) a² + b² = c²

(5)² + (2√14)² = (9)²

25 + (4(14)) = 81

25 + 56 = 81

81 = 81 It IS a right triangle

8.) a² + b² = c²

(6)² + (√49)² = (√82)²

36 + 49 = 82

85 = 82 It is NOT a right triangle

9.) a² + b² = c²

(13)² + (2√39)² = (16)²

169 + (4(39)) = 256

169 + 156 = 256

325 = 256 It is NOT a right triangle

7 0

3 years ago

Graph and label the image of the figure below after a dilation by a factor of 1/2.

neonofarm [45]

Answer:

M' (1.5, -1), F' (2, -1), L' (0.5 -2.5), W' (2.5, -2.5)

see graph below

Explanation:

Given:

The image of a quadrilateral on a coordinate plane

To find:

The coordinates of the new image after dilation of 1/2 have been applied to the original image.

Then graph the coordinates

First, we need to state the coordinates of the original image:

M = (3, -2)

F = (4, -2)

L = (1, -5)

W = (5, -5)

Next, we will apply a scale factor of 1/2:

$\begin{gathered} Dilation\text{ rule:} \\ (x,\text{ y\rparen}\rightarrow(kx,\text{ ky\rparen} \\ where\text{ k = scale factor} \\ \\ scale\text{ factor = 1/2} \\ M^{\prime}\text{ = \lparen}\frac{1}{2}(3),\text{ }\frac{1}{2}(-2)) \\ M^{\prime}\text{ = \lparen}\frac{3}{2},\text{ -1\rparen} \\ \\ F\text{ = \lparen}\frac{1}{2}(4),\text{ }\frac{1}{2}(-2)) \\ F^{\prime}\text{ = \lparen2, -1\rparen} \end{gathered}$ $\begin{gathered} L\text{ = \lparen}\frac{1}{2}(1),\text{ }\frac{1}{2}(-5)) \\ L^{\prime}\text{ = \lparen}\frac{1}{2},\text{ }\frac{-5}{2}) \\ \\ W\text{ = \lparen}\frac{1}{2}(5),\text{ }\frac{1}{2}(-5)) \\ W^{\prime}\text{ = \lparen}\frac{5}{2},\text{ }\frac{-5}{2}) \end{gathered}$

The new coordinates:

M' (3/2, -1), F' (2, -1), L' (1/2, -5/2), W' (5/2, -5/2)

M' (1.5, -1), F' (2, -1), L' (0.5 -2.5), W' (2.5, -2.5)

Plotting the coordinates: