All categories

3 years ago

PLZ HELP ME!!!!!

Mathematics

1 answer:

saveliy_v [14]3 years ago

3 0

5b-2=18
.............

You might be interested in

Find the axis of symmetry, domain, and range for the function f(x)=3(x-22)^2 +6

vekshin1

React to my new music video Dasgasdom3 East Coast junkie

3 0

3 years ago

I need help pls this is really hard

Greeley [361]

Answer:

I cant see it wrong angle lol

8 0

3 years ago

Read 2 more answers

A circle is growing so that the radius is increasing at the rate of 3 cm/min. How fast is the area of the circle changing at the

Naya [18.7K]

Answer:

The area is growing at a rate of $\frac{dA}{dt} =226.2 \,\frac{cm^2}{min}$

Step-by-step explanation:

<em>Notice that this problem requires the use of implicit differentiation in related rates (some some calculus concepts to be understood), and not all middle school students cover such.</em>

We identify that the info given on the increasing rate of the circle's radius is 3 $\frac{cm}{min}$ and we identify such as the following differential rate:

$\frac{dr}{dt} = 3\,\frac{cm}{min}$

Our unknown is the rate at which the area (A) of the circle is growing under these circumstances,that is, we need to find $\frac{dA}{dt}$ .

So we look into a formula for the area (A) of a circle in terms of its radius (r), so as to have a way of connecting both quantities (A and r):

$A=\pi\,r^2$

We now apply the derivative operator with respect to time ( $\frac{d}{dt}$ ) to this equation, and use chain rule as we find the quadratic form of the radius:

$\frac{d}{dt} [A=\pi\,r^2]\\\frac{dA}{dt} =\pi\,*2*r*\frac{dr}{dt}$

Now we replace the known values of the rate at which the radius is growing ( $\frac{dr}{dt} = 3\,\frac{cm}{min}$ ), and also the value of the radius (r = 12 cm) at which we need to find he specific rate of change for the area :

$\frac{dA}{dt} =\pi\,*2*r*\frac{dr}{dt}\\\frac{dA}{dt} =\pi\,*2*(12\,cm)*(3\,\frac{cm}{min}) \\\frac{dA}{dt} =226.19467 \,\frac{cm^2}{min}\\$

which we can round to one decimal place as:

$\frac{dA}{dt} =226.2 \,\frac{cm^2}{min}$

4 0

4 years ago

Given the box plot, will the mean or the median provide a better description of the center? (1 point)

Lady_Fox [76]

The data are bunched towards the higher values above the mean, but spread out to values below the mean. Therefore the median provides a better description of the center.

7 0

3 years ago

Read 2 more answers

The probability density of a random variable X is given in the figure below.

vekshin1

The answer is actually 62

8 0

3 years ago