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2 years ago

Help please .

Mathematics

1 answer:

Firdavs [7]2 years ago

6 0

Answer:

Step-by-step explanation:

(9+9)+(7+7)= 18+14=32

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Is -16/-2 rational or irrational?

mel-nik [20]

Answer:

Totally Rational

Step-by-step explanation:

-16/-2 would give you 8 which is a rational number

I hope this helps, dont hesitate to ask for any question.Mark me as brainliest is appreciated.Tq!!

4 0

3 years ago

How do you express height in feet by only using a mixed number

guajiro [1.7K]

Height and then the inches over 12.
Ex. 5'4" = 5 4/12 or 5 1/3

8 0

3 years ago

In the adjoining figure, ABCD is a square of side 28 cm find the area of shaded region

vodka [1.7K]

Step-by-step explanation:

Area of square

A = s²

A = 28²

A = 28 × 28

A = 784 cm²

7 0

2 years ago

Could I get some help with this question please?

icang [17]

Opposite
additive inverse

5 0

3 years ago

The graphs of the polar curves r = 4 and r = 3 + 2cosθ are shown in the figure above. The curves intersect at θ = π/3 and θ = 5π

Gennadij [26K]

(a)

$\displaystyle \frac{1}{2} \cdot \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \left(4^2 - (3 + 2\cos\theta)^2 \right) \, d\theta$

or, via symmetry

$\displaystyle\frac{1}{2} \cdot 2 \int_{\frac{\pi}{3}}^{\pi} \left(4^2 - (3 + 2\cos\theta)^2 \right) \, d\theta$

____________

(b)

By the chain rule:

$\displaystyle \frac{dy}{dx} = \frac{ dy/ d\theta}{ dx/ d\theta}$

For polar coordinates, x = rcosθ and y = rsinθ. Since
r = 3 + 2cosθ, it follows that

$x = (3 + 2\cos\theta) \cos \theta \\ 
y = (3 + 2\cos\theta) \sin \theta$

Differentiating with respect to theta

$\begin{aligned}
\displaystyle \frac{dy}{dx} &= \frac{ dy/ d\theta}{ dx/ d\theta} \\
&= \frac{(3 + 2\cos\theta)(\cos\theta) + (-2\sin\theta)(\sin\theta)}{(3 + 2\cos\theta)(-\sin\theta) + (-2\sin\theta)(\cos\theta)} \\ \\
\left.\frac{dy}{dx}\right_{\theta = \frac{\pi}{2}}
&= 2/3
\end{aligned}$

2/3 is the slope

____________

(c)

"distance between the particle and the origin increases at a constant rate of 3 units per second" implies dr/dt = 3

Angle θ and r are related via r = 3 + 2cosθ, so implicitly differentiating with respect to time

 $\displaystyle\frac{dr}{dt} = -2\sin\theta \frac{d\theta}{dt} \quad \stackrel{\theta = \pi/3}{\implies} \quad 3 = -2\left( \frac{\sqrt{3}}{2}}\right) \frac{d\theta}{dt} \implies \\ \\ \frac{d\theta}{dt} = -\sqrt{3} \text{ radians per second}$

5 0

3 years ago