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2 years ago

Which graph shape represents the polar curve r = 1 + 4sin(θ)?

Mathematics

2 answers:

umka2103 [35]2 years ago

7 0

Answer:

A cardioid

Step-by-step explanation:

A cardioid is a curve in the shape of a heart traced by a point on the circumference of a circle as it rolls around a fixed circle of the same radius.

The equation of vertical cardioid is $r=a$ ± $a\sin \theta$

Given the polar curve is $r=$ $1+$ 4 $\sin \theta$

This is an equation of vertical cardioid.

So, it represents a cardioid.

Tomtit [17]2 years ago

6 0

Answer:

C - a limaçon with an inner loop

Step-by-step explanation:

If you have a graphing calculator, use it to plug in the equation.

Based on what you have learned, you will recognize the graph is C

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Ksivusya [100]

Answer:

-85, -1.4, -0.9, 0.6, 14

Step-by-step explanation:

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3 years ago

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NemiM [27]

Answer:

V = 235.6 cm³

Step-by-step explanation:

the formula for the volume of a cylinder of radius r and length h is

V = πr²h.

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V = π(2.5 cm)²(12 cm) = 75π cm³

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3 years ago

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Multiply or divide as indicated. <br><br> x 2 • x 5

WARRIOR [948]

Hi there!

${x}^{2} \times {x}^{5} = x {}^{2 + 5} = {x}^{7}$

We can also write is out entirely.
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${x}^{2} \times {x}^{5} = \\ (x \times x ) \times (x \times x \times x \times x \times x) = \\ x \times x \times x \times x \times x \times x \times x = \\ {x}^{7}$

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3 years ago

Need help bad !!!!!!!!!!!!

Nonamiya [84]

There are 2 order pairs there (6,-4),(-2,4)
find the midpoint using the formula

(x1+x2)/2 , (y1+y2)/2
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Ivahew [28]

The rolls of the dice are independent, i.e. the outcome of the second die doesn't depend in any way on the outcome of the first die.

In cases like this, the probability of two events happening one after the other is the multiplication of the probabilities of the two events.

So, the probability of rolling two 6s is the multiplication of the probabilities of rolling a six with the first die, and another six with the second:

$P(\text{rolling two 6s}) = P(\text{rolling a 6}) \cdot P(\text{rolling a 6}) = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$

Similarly,

$P(\text{rolling two 3s}) = P(\text{rolling a 3}) \cdot P(\text{rolling a 3}) = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$

Actually, you can see that the probability of rolling any ordered couple is always 1/36, since the probability of rolling any number on both dice is 1/6:

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