I would like to ask about question d

Suppose that circles A and B have a central angle measuring 100°. Additionally, the measure of the sector for circle A is 10π m2

lesya692 [45]

We know that
When circles have the same central angle measure, the ratio of measure of the sectors is the same as the ratio of the radii squared.
so
rA²/rB²=A/B
where
rA-------> radius circle A
rB------> radius circle B
A------> area sector A
B------> area sector B
rB²=B*rA²/A-------> 6²*[40π]/[10π]-------> 144--------> rB=√144
rB=12 m

the answer is
radius of circle B is 12 m

5 0

3 years ago

Read 2 more answers

I NEED AN ANSWER NOW!!! This is passed due!!! Dhruv is a car salesman. He is paid a salary of $2,200 per month, plus $500 for ea

klio [65]

Answer:

B.
D: {0, 1, 2, 3,...}
R: {$2200, $2700, $3200, $3700,...}

Step-by-step explanation:

Given function is

f(x) = 500x + 2200

The graph of this function is attached

Domain is input value
Range is output value

Correct set of domains and ranges of the given function is:

Option B
D: {0, 1, 2, 3,...}
R: {$2200, $2700, $3200, $3700,...}

8 0

3 years ago

What is the derivative of arctan (y/x) with respect to y??

ycow [4]

arctan(y/x) = d/dx
=1/x∗1/(1+(y/x)^2)

8 0

3 years ago

Give the most accurate name for this figure.

Murrr4er [49]

Answer:

That is a quadrilateral good sir

-edge 2021

4 0

3 years ago

At a price of $p the demand x per month (in multiples of 100) for a new piece of software is given by x 2 + 2xp + 4p 2 = 5200. B

WINSTONCH [101]

Answer:

The rate of decrease in demand for the software when the software costs $10 is -100

Step-by-step explanation:

Given the function of price $p the demand x per month as,

$x^{2}+2xp+4p^{2}=5200$

Also given that, the price is increasing at the rate of 70 dollar per month.

$\therefore \dfrac{dp}{dt}=70$ .

To find rate of decrease in demand, differentiate the given function with respect to t as follows,

$\dfrac{d}{dt}\left(x^2+2xp+4p^2\right)=\dfrac{d}{dt}\left(5200\right)$

Applying sum rule and constant rule of derivative,

$\dfrac{d}{dt}\left(x^2\right)+\dfrac{d}{dt}\left(2xp\right)+\dfrac{d}{dt}\left(4p^2\right)=0$

Applying constant multiple rule of derivative,

$\dfrac{d}{dt}\left(x^2\right)+2\dfrac{d}{dt}\left(xp\right)+4\dfrac{d}{dt}\left(p^2\right)=0$

Applying power rule and product rule of derivative,

$2x^{2-1}\dfrac{dx}{dt}+2\left(x\dfrac{dp}{dt}+p\dfrac{dx}{dt}\right)+4\left(2p^{2-1}\right)\dfrac{dp}{dt}=0$

Simplifying,

$2x\dfrac{dx}{dt}+2\left(x\dfrac{dp}{dt}+p\dfrac{dx}{dt}\right)+8p\dfrac{dp}{dt}=0$

Now to find the value of x, substitute the value of p=$10 in given equation.

$x^{2}+2x\left(10\right)+4\left(10\right)^{2}=5200$

$x^{2}+20x+400=5200$

Subtracting 5200 from both sides,

$x^{2}+20x+400-5200=0$

$x^{2}+20x-4800=0$

To find the value of x, split the middle terms such that product of two number is 4800 and whose difference is 20.

Therefore the numbers are 80 and -60.

$x^{2}+80x-60x-4800=0$

Now factor out x from $x^{2}+80x$ and 60 from $60x-4800$

$x\left(x+80\right)-60\left(x+80\right)=0$

Factor out common term x+80,

$\left(x+80\right)\left(x-60\right)=0$

By using zero factor principle,

$\left(x+80\right)=0$ and $\left(x-60\right)=0$

$x=-80$ and $x=60$

Since demand x can never be negative, so x = 60.

Now,

$2x\dfrac{dx}{dt}+2\left(x\dfrac{dp}{dt}+p\dfrac{dx}{dt}\right)+8p\dfrac{dp}{dt}=0$

Substituting the value.

$2\left(60\right)\dfrac{dx}{dt}+2\left(60\left(70\right)+10\dfrac{dx}{dt}\right)+8\left(10\right)\left(70\right)=0$

Simplifying,

$120\dfrac{dx}{dt}+2\left(4200+10\dfrac{dx}{dt}\right)+5600=0$

$120\dfrac{dx}{dt}+8400+20\dfrac{dx}{dt}+5600=0$

Combining common term,

$140\dfrac{dx}{dt}+14000=0$

Subtracting 14000 from both sides,

$140\dfrac{dx}{dt}=-14000$

Dividing 140 from both sides,

$\dfrac{dx}{dt}=-\dfrac{14000}{140}$

$\dfrac{dx}{dt}=-100$

Negative sign indicates that rate is decreasing.

Therefore, the rate of decrease in demand of software is -100

6 0

3 years ago