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3 years ago

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Mathematics

1 answer:

frutty [35]3 years ago

3 0

Answer:

The answer is B

Step-by-step explanation:

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Evaluate 3z when z=9

sweet-ann [11.9K]

Z=9 so 3*9 is 27. 27 is your answer. You are literally just switching the variable z with 9.

3 0

3 years ago

Read 2 more answers

Samantha wants to use her savings of 1150 to buy shirts and watches for her family. The total price of the shirt she bought was

ioda

Answer-

The maximum number of watches that Samantha come by with her savings is 10.

Solution-

The amount of money Samantha has in her savings account = $1150

She wants to buy shirts and watches.

Cost of one shirt = $84

Cost of each watch = $99

Let she can buy maximum of x watches, so the net price of the watches is $99x.

Then,

$\Rightarrow 99x+84=1150\\\\\Rightarrow 99x=1150-84\\\\\Rightarrow 99x=1066\\\\\Rightarrow x=\dfrac{1066}{99}\\\\\Rightarrow x=10.77$

As the number of watches can not be in fraction, so at most she can buy 10 watches.

8 0

4 years ago

Integrate this for me please

mel-nik [20]

$\int\limits_{0}^{\frac{\pi }{6}}tan^2(x)sec^2(x)\cdot dx
\\------------------\\
u=tan(x)\implies \frac{du}{dx}=sec^2(x)\implies \frac{du}{sec^2(x)}=dx
\\------------------\\
\int\limits_{0}^{\frac{\pi }{6}}u^2sec^2(x)\cdot \cfrac{du}{sec^2(x)}\implies 
\int\limits_{0}^{\frac{\pi }{6}}u^2\cdot du
\\ \quad \\
$

$\textit{now, we need to change the bounds as well, so}
\\------------------\\
u(0)=tan(0)\implies 0
\\ \quad \\

u\left( \frac{\pi }{6} \right)=tan\left( \frac{\pi }{6} \right)\implies \frac{1}{\sqrt{3}}
\\------------------\\
thus\implies \int\limits_{0}^{\frac{1 }{\sqrt{3}}}u^2\cdot du$

and surely you can take it from there,
recall, that, since we changed the bounds, with the u(x),
you don't need to change the variable "u", and simply,
get the integral of it, simple enough, and apply those bounds

5 0

3 years ago

Read 2 more answers

Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the abo

Likurg_2 [28]

Given :

Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the above.

To Find :

At what rate is the height of the cone increasing when the height is 2 cm form the base of the cone.

Solution :

We know, volume of a cone is given by :

$V = \dfrac{\pi r^2h}{3}$

Also, h = r/5

r = 5h

$V = \dfrac{\pi (5h)^2h}{3}\\\\V = \dfrac{\pi 25h^3}{3}$

Differentiating above equation w.r.t h, we get :

$\dfrac{dV}{dt} = \dfrac{d}{dt}(\dfrac{\pi 25h^3}{3})\\\\\dfrac{dV}{dh} = 25\pi h^2\dfrac{dh}{dt}\\\\\dfrac{dh}{dt}= \dfrac{\dfrac{dV}{dh}}{25\pi h^2 }\\\\\dfrac{dh}{dt}=\dfrac{10}{25\times 3.14\times 2^2}\\\\\dfrac{dh}{dt}=0.032\ cm/s$

Hence, this is the required solution.

5 0

3 years ago

Urgent Find the value of x

kipiarov [429]

Answer:

Step-by-step explanation:

6 0

4 years ago