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1 year ago

Which arrangement shows −275 , −4.1 , −535 , and −3 in order from least to greatest?

Mathematics

1 answer:

vodomira [7]1 year ago

6 0

Answer:

-535, -275, -4.1, -3

Step-by-step explanation:

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Eliot needs to replace a window that is in the shape of a semicircle.

USPshnik [31]

Answer:

////

Step-by-step explanation:

6 0

1 year ago

If f(x)=x2+10sinx, show that there is a number c such that f(c)=1000.

gayaneshka [121]

F(x) is continuous for all x.

Pick a point and show that f(x) is either negative or positive. Pick another point and show that f(x) is negative, if positive, or positive, if negative.

At x = 30, f(30) - 1000 = 900 + 10sin(30) - 1000 ≤ 0
Now, show at another point f(x) - 1000 is positive, and hence, there would be root between 30 and such point.

Let's pick 40.
At x = 40, f(40) - 1000 = 1600 + 10sin(40) - 1000 ≥ 0

Since f(x) - 1000 is continuous, there lies a root between 30 and 40, and hence, 30 ≤ c ≤ 40

7 0

2 years ago

The radius of a right circular cone is increasing at a rate of 1.4 in/s while its height is decreasing at a rate of 2.1 in/s. At

kow [346]

Given:

$\dfrac{dr}{dt}=1.4\text{ in/s}$

$\dfrac{dh}{dt}=-2.1\text{ in/s}$

To find:

The rate of change in volume at $r=120\text{ in. and }175\text{ in.}$

Solution:

We know that, volume of a cone is

$V=\dfrac{1}{3}\pi r^2h$

Differentiate with respect to t.

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

Substitute the given values.

$\dfrac{dV}{dt}=\dfrac{1}{3}\times \dfrac{22}{7}\times \left[(120)^2(-2.1) +175(2)(120)(1.4)\right]$

$\dfrac{dV}{dt}=\dfrac{22}{21}\times \left[-30240+58800\right]$

$\dfrac{dV}{dt}=\dfrac{22}{21}\times 28560$

$\dfrac{dV}{dt}=29920$

Therefore, the volume of decreased by 29920 cubic inches per second.

6 0

1 year ago

Jack works as a waiter and is keeping track of the tips he earns daily. About how much does Jack have to earn in tips on

Ray Of Light [21]

Answer:

25$

Step-by-step explanation:

Add all of the days up and divide by 7

7 0

2 years ago

The spring has a stiffness k=200n/m and is unstretched when the 25 kg block is at

g100num [7]

To solve this we are going to use the formula fro the force applied to a spring: $F=ks$
where
$k$ is the spring constant
$s$ is the extension

Since we know the $F=ma$ , we can replace that in our formula and solve for $a$ :
$ma=ks$
$a= \frac{ks}{m}$
where
$a$ is the acceleration
$k$ is the spring constant
$s$ is the extension
$m$ is the mass

We know for our problem that $k=200$ , $s=0.4$ , and $m=25$ . So lets replace those values in our formula to find $a$ :
$a= \frac{ks}{m}$
$a= \frac{(200)(0.4)}{25}$
$a=3.5 \frac{m}{s^2}$

We can conclude that the acceleration of the block when s=0.4m is $3.5 \frac{m}{s^2}$ .

7 0

1 year ago