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3 years ago

Help me plz idk how

Mathematics

1 answer:

Alika [10]3 years ago

8 0

Hello there,

For problem one
The mile is an appropriate unit to measure the distance run over a one-week period.

For problem two
The pound is an appropriate unit to measure the weight of a bag of sugar.

For problem three
The yard is an appropriate unit to measure the distance run on a football field.

Hope this helps!

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A park is 5 miles east of Roxana's home. A library is 4 miles north of the park. How far is Roxana's home from the library?

Ber [7]

Answer: 9 miles I think but I dont know where her house is exactly.

8 0

3 years ago

Write equation for the center(-1,0) radius of 6

Ira Lisetskai [31]

Answer:

(x+1)² + y² = 36

Step-by-step explanation:

Equation of a circle:

(x-h)² + (y-k)² = r²

Center: (h,k)

Radius: r

Given a center of (-1,0) and a radius of 6, we have for our equation:

(x-(-1))² + (y-0)² = 6²

(x+1)² + y² = 36

3 0

3 years ago

13/15 + 1/8 = Please help

kogti [31]

Reduce the fraction using 3
13/15+1/8

13/15=1/5

1/5+1/8

8+5/40

fraction response: 13/40
answer in decimal numbers: 0.325

3 0

3 years ago

Read 2 more answers

help plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz plz

Rudiy27

Answer:

Q. 4.

-> $3,333

Q. 5

-> $40,000

3 0

3 years ago

Read 2 more answers

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