All categories

3 years ago

Is 7.401 larger or smaller then 7.04

Mathematics

2 answers:

tester [92]3 years ago

6 0

Answer:

Larger.

7.401 = 7.40 (rounding)

7.40 > 7.04

Bye!

rosijanka [135]3 years ago

5 0

Answer:

Larger

Step-by-step explanation:

Since, 7.401 > 7.04

Therefore, 7.401 is larger than 7.04.

You might be interested in

PLEASES HELP<br> I need help on this question the screenshot is down below

hoa [83]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

For which value of x does f(x) = g(x)

Natali5045456 [20]

Answer:

x=-3 and x=0

Step-by-step explanation:

The values where f(x) = g(x) is where the graphs intersect

One point is at (-3,0) and the other point is at (0,6)

We have two points of intersection since one is a parabola and the other is a line

8 0

3 years ago

Yesenia took three tests in her accounting class. She scored 91 points on her first test and 73 points on her second test. If he

Dmitry_Shevchenko [17]

The answer is 73
you just set it up at 73+91+x=79 and solve

3 0

3 years ago

Read 2 more answers

HELP ASAP WITH MATH DECIMALS!!! I WILL UPROVE!!!

andriy [413]

Total time taken to attach the wheels and hood to a car = 8.216 + 3.652 = 11.868 seconds

Total time taken to attach good and wheels to 425 cars = 11.868 x 425 = 4925.22 seconds

Ans. 4925.22 seconds

8 0

3 years ago

a rectangle is constructed with its base on the x-axis and two of its vertices on the parabola y=121-x^2. what are the dimension

insens350 [35]

Answer:

the length is $\dfrac{22}{\sqrt{3}}$ and the width is $121-\left(\dfrac{11}{\sqrt{3}}\right)^2=121-\dfrac{121}{3}=\dfrac{242}{3}.$

Step-by-step explanation:

Let points A and B be placed on the x-axis. Their coordinates are $A(x_0,0)\ (x_0>0)$ and $B(-x_0,0)$ (because of parabola symmetry). Two other vertices lie on the parabola, then $C(-x_0,121-x_0^2)$ and $D(x_0,121-x_0^2).$ The length of the side AB is $2x_0$ and the length of the side AD is $121-x_0^2.$ Thus, the area of the rectangle ABCD is

$A=2x_0\cdot (121-x_0^2)=242x_0-2x_0^3.$

Find the derivative A':

$A'=242-2\cdot 3x_0^2=242-6x_0^2.$

Equate A' to 0:

$242-6x_0^2=0,\\ \\x_0^2=\dfrac{121}{3},\\ \\x_0=\dfrac{11}{\sqrt{3}}.$

The maximum area of the rectangle is

$A_{max}=242\cdot \dfrac{11}{\sqrt{3}}-2\cdot \left(\dfrac{11}{\sqrt{3}}\right)^3=\dfrac{2662}{\sqrt{3}}-\dfrac{2662}{3\sqrt{3}}=\dfrac{5324}{3\sqrt{3}}\ un^2.$

The dimensions of the rectangle are:

the length is $\dfrac{22}{\sqrt{3}}\ un.$ and the width is $121-\left(\dfrac{11}{\sqrt{3}}\right)^2=121-\dfrac{121}{3}=\dfrac{242}{3}\ un.$

6 0

4 years ago

Read 2 more answers