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3 years ago

Conjecture: The difference between two negative numbers is always negative. Write a counterexample. *

Mathematics

1 answer:

Katen [24]3 years ago

4 0

If you add two negative numbers you get a sum that is less than both of the numbers. so -1+-2=-3

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The radius of a right circular cone is increasing at a rate of 1.9 in/s while its height is decreasing at a rate of 2.2 in/s. At

musickatia [10]

Answer:

Volume of the cone is increasing at the rate $9916\pi \frac{in^3}{s}$ .

Step-by-step explanation:

Given: The radius of a right circular cone is increasing at a rate of $1.9$ in/s while its height is decreasing at a rate of $2.2$ in/s.

To find: The rate at which volume of the cone changing when the radius is $134$ in. and the height is $136$ in.

Solution:

We have,

$\frac{dr}{dt} =1.9 \:\text{in/s}$ , $\frac{dh}{dt}=-2.2\:\text{in/s}$ , $r=134 \:\text{in}$ , $h=136\:\text{in}$

Now, let $V$ be the volume of the cone.

So, $V=\frac{1}{3}\pi r^{2}h$

Differentiate with respect to $t$ .

$\frac{dv}{dt} =\frac{1}{3}\pi \left [ r^2\frac{dh}{dt}+h\left ( 2r \right )\frac{dr}{dt} \right ]$

Now, on substituting the values, we get

$\frac{dv}{dt} =\frac{1}{3}\pi\left [ \left ( 134 \right )^2\left ( -2.2 \right )+\left ( 136\right )\left ( 2 \right )\left ( 134 \right )\left ( 1.9 \right ) \right ]$

$\frac{dv}{dt} =\frac{1}{3}\pi\left [ -39503.2+69251.2 \right ]$

$\frac{dv}{dt} =\frac{1}{3}\pi\left [ 29748 \right ]$

$\frac{dv}{dt} =9916\pi \frac{in^3}{s}$

Hence, the volume of the cone is increasing at the rate $9916\pi \frac{in^3}{s}$ .

6 0

3 years ago

What is the circumference of a circle with a radius of 2.5 m?

mixer [17]

The circumference of a circle is always calculated by 2 $\pi$ R
So if R=2.5 m this circumference is 2* $\pi$ *2.5 m
Which is 5 $\pi$ m (choice B)

Hope this helped!

4 0

3 years ago

Read 2 more answers

I need help with this Compound Inequality.

Liula [17]

I don’t know if this is right but i looked it up because i was clueless… it says, “(-1,29) ∪ (26,50) (-3,5) ∪ (5,10) (-1,2) ∪ (-4,0)

6 0

3 years ago

1+secA/sec A = sin^2 A / 1-cos A

Fofino [41]

Answer: see proof below

<u>Step-by-step explanation:</u>

$\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos A}$

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

<u>Proof LHS → RHS</u>

$\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}$

$\text{Identity:}\qquad \qquad \dfrac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}$

$\text{Simplify:}\qquad \qquad \dfrac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\\\\\.\qquad \qquad \qquad =\dfrac{1+\cos A}{1}$

$\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}$

$\text{Identity:}\qquad \qquad \dfrac{\sin^2 A}{1-\cos A}$

$\text{LHS = RHS:}\quad \dfrac{\sin^2 A}{1-\cos A}=\dfrac{\sin^2 A}{1-\cos A}\quad \checkmark$

3 0

4 years ago

Write an equation for the line that is parallel to the given line and passes through the

artcher [175]

I hope this helps you

4 0

4 years ago

Read 2 more answers