All categories

3 years ago

I need.help please

Mathematics

2 answers:

Scrat [10]3 years ago

8 0

It is 100 times greater

EleoNora [17]3 years ago

3 0

The digit 5 in the thousands place is 4950 greater than the digit 5 in the tens place.

You might be interested in

Is y= 3(1/2)^x increasing or decreasing?

Zigmanuir [339]

Answer: Hey I looked on a graphing calculator on Desmos, and its seems that this is increasing.

Step-by-step explanation:

5 0

2 years ago

HELP PLEASEEE ASAP ! THANK YOU :)

Salsk061 [2.6K]

Answer:

f(x) = 2x³ - 6x² - 48x + 24

relative minimum: (4,-136)

relative maximum: (-2,80)

answer is the second option

6 0

3 years ago

Read 2 more answers

Which equation is equivalent to

Paul [167]

Answer:

c) 3x - 45 = 10

Step-by-step explanation:

$\frac{1}{5}x-3=\frac{2}{3}\\\\\frac{1}{5}x-\frac{3*5}{1*5}=\frac{2}{3}\\\\\frac{x-15}{5}=\frac{2}{3}\\$

Cross multiply,

3*(x - 15) = 2*5

3*x - 3*15 = 10

3x - 45 = 10

4 0

3 years ago

The length of the box is 2x - 2 and the width is x - 5. Find the perimeter. (HINT: p= 2L + 2W) Choose one.

trasher [3.6K]

Answer:

Option D: 6x-14 is the correct answer.

Step-by-step explanation:

Given that:

Length of the box = L = 2x-2

Width of the box = W = x-5

Perimeter is defined by the sum of length of all sides of the box.

Perimeter of the box = P = 2L + 2W

Perimeter of box = 2(2x-2) + 2(x-5)

Perimeter = 4x-4+2x-10

Perimeter = 6x-14

The perimeter of the box is 6x-14.

Hence,

Option D: 6x-14 is the correct answer.

6 0

3 years ago

A tank has the shape of a surface generated by revolving the parabolic segment y = x2 for 0 ≤ x ≤ 3 about the y-axis (measuremen

Darina [25.2K]

Answer:

$100\pi\int\limits^9_0 {(\sqrt y)^2(14-y)} \, dy$ ft-lbs.

Step-by-step explanation:

Given:

The shape of the tank is obtained by revolving $y=x^2$ about y axis in the interval $0\leq x\leq 3$ .

Density of the fluid in the tank, $D=100\ lbs/ft^3$

Let the initial height of the fluid be 'y' feet from the bottom.

The bottom of the tank is, $y(0)=0^2=0$

Now, the height has to be raised to a height 5 feet above the top of the tank.

The height of top of the tank is obtained by plugging in $x=3$ in the parabolic equation . This gives,

$H=3^2=9\ ft$

So, the height of top of tank is, $y(3)=H=9\ ft$

Now, 5 ft above 'H' means $H+5=9+5=14$

Therefore, the increase in height of the top surface of the fluid in the tank is given as:

$\Delta y=(14-y)$ ft

Now, area of cross section of the tank is given as:

$A(y)=\pi r^2\\r\to radius\ of\ the\ cross\ section$

Radius is the distance of a point on the parabola from the y axis. This is nothing but the x-coordinate of the point.

We have, $y=x^2$

So, $x=\sqrt y$

Therefore, radius, $r=\sqrt y$

Now, area of cross section is, $A(y)=\pi (\sqrt y)^2$

Work done in pumping the contents to 5 feet above is given as:

$W=D\int\limits^{y(3)}_{y(0)} {A(y)(\Delta y)} \, dy$

Plug in all the values. This gives,

$W=100\int\limits^9_0 {\pi (\sqrt y)^2(14-y)} \, dy\\\\W=100\pi\int\limits^9_0 { (\sqrt y)^2(14-y)} \, dy\textrm{ ft-lbs}$

7 0

2 years ago