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2 years ago

Use your grapher to determine which of the graphs matches the polar equation r = 3 sin 2θ.

Mathematics

1 answer:

VikaD [51]2 years ago

4 0

Given:

r = 3 sin 2θ 
Find: Graph that matches the polar equation Solution:

Use the polar graph type in the graphing calculator and enter the following values: 3sin (2 theta). Please see the attached image for the answer. The graph would look like butterfly wings.

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Choose the algebraic expression that represents the area of a triangle that has a base 8 inches less than 2 times wider than the

loris [4]

Answer:

B. (1/2)(2h-8)(h)

Step-by-step explanation:

Area of a triangle = 1/2 × base × height

Let

Height = h

Base = 8 inches less than 2 times wider than the hight

= 2h-8

Substituting the values of base and height into the formula

We have,

Area of a triangle = 1/2 × base × height

= 1/2 × (2h-8) × h

= (1/2)(2h-8)(h)

The correct answer is option

B. (1/2)(2h-8)(h)

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2 years ago

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Y = 12x - 40

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2 years ago

A circle has a circumference of 153.86153.86153, point, 86 units.

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2 years ago

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Which of the following functions is graphed below?

dalvyx [7]

Answer:

The function graphed is y = Ix - 5I - 4 ⇒ answer D

Step-by-step explanation:

* Lets explain how to solve this problem

- From the graph

# The graph intersects the x-axis at x = 1 and x = 9

∴ The x-intercepts are 1 and 9

- We can find x-intercept by equate y by 0

* Lets equate the answers by zero to find the x-intercept

# y = Ix + 5I - 4

∵ Ix + 5I - 4 = 0

- Add 4 to both sides

∴ Ix + 5I = 4

- Remember in Ia + bI = c, then we have two answers :

a + b = c OR a + b = -c

∵ Ix + 5I = 4

∴ x + 5 = 4 ⇒ subtract 5 from both sides

∴ x = -1

- OR

∴ x + 5 = -4 ⇒ subtract 5 from both sides

∴ x = -9

∴ The x-intercepts are -1 and -9 not the same with figure

# y = Ix - 5I + 4

∵ Ix - 5I + 4 = 0

- Subtract 4 from both sides

∴ Ix - 5I = -4

- Remember in Ia + bI = c , c can't be negative because the absolute

value is always positive

∴ We can't solve this equation

# y = Ix + 5I + 4

∵ Ix + 5I + 4 = 0

- Subtract 4 from both sides

∴ Ix + 5I = -4

- Remember in Ia + bI = c , c can't be negative because the absolute

value is always positive

∴ We can't solve this equation

# y = Ix - 5I - 4

∵ Ix - 5I - 4 = 0

- Add 4 to both sides

∴ Ix - 5I = 4

- Remember in Ia + bI = c, then we have two answers :

a + b = c OR a + b = -c

∵ Ix - 5I = 4

∴ x - 5 = 4 ⇒ add 5 to both sides

∴ x = 9

- OR

∴ x - 5 = -4 ⇒ add 5 to both sides

∴ x = 1

∴ The x-intercepts are 1 and 9 the same with figure

* The function graphed is y = Ix - 5I - 4

5 0

2 years ago