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VMariaS [17]
3 years ago
6

Which statement is true about the value of |-14|?

Mathematics
1 answer:
kondor19780726 [428]3 years ago
6 0
It is the distance between -14 and 0 on the number line
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The local kennel has 3 spots for cats to every 5 dogs. if the local kennel has a total of 72 spots for both cats and dogs, then
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3:5
5+3= 8
72 / 8 = 9
9 x 3 = 27
27 spots for cats
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En la fábrica de Don Toño hoy fabricaron 320 pantalones, si los guardan en cajas donde caben 50 ¿cuántas cajas llenaron?
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6.4 Cajas

Step-by-step explanation:

320/50 = 6.4 Cajas

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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
<span>r = 3 + 2cosθ, it follows that

x = (3 + 2\cos\theta) \cos \theta \\ &#10;y = (3 + 2\cos\theta) \sin \theta

Differentiating with respect to theta

\begin{aligned}&#10;\displaystyle \frac{dy}{dx} &= \frac{ dy/ d\theta}{ dx/ d\theta} \\&#10;&= \frac{(3 + 2\cos\theta)(\cos\theta) + (-2\sin\theta)(\sin\theta)}{(3 + 2\cos\theta)(-\sin\theta) + (-2\sin\theta)(\cos\theta)} \\ \\&#10;\left.\frac{dy}{dx}\right_{\theta = \frac{\pi}{2}}&#10;&= 2/3&#10;\end{aligned}

2/3 is the slope

____________

(c)

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

A</span>ngle θ and r are related via <span>r = 3 + 2cosθ, so implicitly differentiating with respect to time

</span><span />\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
Type in the correct value for m... (Only the numerical answer)<br> 369 + m = 908
Hunter-Best [27]
Answer

M = 539

Expo nation
4 0
3 years ago
I really need help! This is last minute and its 1 am. I'm tired so I'm going to leave it to you guys to solve my problems and pr
Illusion [34]

Answer:

<h2><u>question 1</u></h2>

n^{2} - 20n -96 = 0

use product and sum method

product = -96

sum = -20

numbers needed = ( -24 , 4)

n - 24 = 0

n + 4 = 0

hence <u>n = 24 and n = -4 </u>

<u></u>

<h2><u>Question 2 </u></h2>

<u />x^{2} + 12 x = 48<u />

in the form ax^{2}  +bx +c = 0

= x^{2} +12x - 48

make use of the formula :

\frac{-b+-\sqrt{b^{2} -4ac} }{2a}

replace values to make 2 equations :

1.\frac{-12+\sqrt{12^{2} -4*1*-48} }{2*1} = 3.17

2.\frac{-12-\sqrt{12^{2} -4*1*-48} }{2*1} = -15.2

hence <u>x = 3.17 and x = -15.2</u>

<u />

<h2><u>Question 3 </u></h2>

<u />x^{2} -14x+40=0<u />

use product and sum method

product = 40

sum = -14

numbers needed = (-10 , -4)

x - 10 = 0

x - 4 = 0

hence<u> x = 10 and x = 4</u>

<u />

<h2><u>Question 4 </u></h2>

<u />5b^{2} -20b-18 = 7<u />

in the form ax^{2}  +bx +c = 0

this becomes 5b^{2} -20b-18-7

= 5b^{2} -20b-25

can simplify by 5

= b^{2} -4b-5 =0\\

use product and sum method

product = -5

sum = -4

numbers needed (-5 , 1)

b-5 = 0

b + 1 = 0

hence <u>b = 5 and b = -1</u>

5 0
3 years ago
Read 2 more answers
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