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9966 [12]
3 years ago
13

If (-3, y) lies on the graph of y = (1/2)^x, then y =

Mathematics
1 answer:
labwork [276]3 years ago
6 0

\text{If (-3, y) lies on the graph of}\ \ y = \left(\dfrac{1}{2}\right)^x, \text{then substitute x = -3}\\\text{ to the equation and calculate y}:\\\\y=\left(\dfrac{1}{2}\right)^{-3}=(2)^3=8\\\\Answer:\ \boxed{y=8}\\\\Used:\ a^{-n}=\dfrac{1}{a^n}

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Angle abd measures (4x+10). angle acd measures (5x-2). what is the measure of arc ad
user100 [1]

Answer:116

Step-by-step explanation: 5x-2=4x+10

x=12

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8 0
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What is the equation of the line that haas the x-intercept of -5?
Alexus [3.1K]

Answer:

A

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What is the answer to 16x + 10-10x=2x+14
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Given the set of vertices, determine whether parallelogram ABCD is a rhombus, rectangle or square. List all that apply. A(7,-4),
Sloan [31]

Given:

Vertices of a parallelogram ABCD are A(7,-4), B(-1,-4), C(-1,-12), D(7, -12).

To find:

Whether the parallelogram ABCD is a rhombus, rectangle or square.

Solution:

Distance formula:

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

Using distance formula, we get

AB=\sqrt{(-4-(-4))^2+(-1-7)^2}

AB=\sqrt{(-4+4)^2+(-8)^2}

AB=\sqrt{0+64}

AB=8

Similarly,

BC=\sqrt{(-1-(-1))^2+(12-(-4))^2}=8

CD=\sqrt{(7-(-1))^2+(-12-(-12))^2}=8

AD=\sqrt{(7-7)^2+(-12-(-4))^2}=8

All sides of parallelogram are equal.

AC=\sqrt{(-1-7)^2+(-12-(-4))^2}=8\sqrt{2}

BD=\sqrt{(7-(-1))^2+(-12-(-4))^2}=8\sqrt{2}

Both diagonals are equal.

Since, all sides are equal and both diagonals are equal, therefore, the parallelogram ABCD is a square.

We know that, a square is special case of rectangles and rhombus.

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7 0
3 years ago
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