Ask question

Ask question

All categories

3 years ago

11

What is the value of 4 1/2 - 2 3/4

1 answer:

Ivenika [448]3 years ago

3 0

Answer: 1 3/4

Step-by-step explanation: First, convert 4 1/2 and 2 3/4 to have like denominators.

4 1/2 = 4 2/4 because 1 x 2 = 2 and 2 x 2 = 4 = 2/4

2 3/4 stays the same.

Now, subtract 4 2/4 - 2 3/4 (4 x 4 = 16 + 2 = 18) and (2 x 4 = 8 + 3 = 11)

18/4 - 11/4 = 7/4 or 1 3/4

You might be interested in

solmaris [256]

Can you post this and provide a picture please so I can help you ?

Explanation:

8 0

3 years ago

Write a function using function notation to describe each situation. Find a reasonable domain and range for each function.

sergejj [24]

F(t)= 7.00 x t + 55 A reasonable domain is (0,1,2,3). The range is ($55,$62,$69,$76)

5 0

3 years ago

The boundary of a lamina consists of the semicircles y = 1 − x2 and y = 16 − x2 together with the portions of the x-axis that jo

oksano4ka [1.4K]

Answer:

Required center of mass $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

Step-by-step explanation:

Given semcircles are,

$y=\sqrt{1-x^2}, y=\sqrt{16-x^2}$ whose radious are 1 and 4 respectively.

To find center of mass, $(\bar{x},\bar{y})$ , let density at any point is $\rho$ and distance from the origin is r be such that,

$\rho=\frac{k}{r}$ where k is a constant.

Mass of the lamina=m= $\int\int_{D}\rho dA$ where A is the total region and D is curves.

then,

$m=\int\int_{D}\rho dA=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}rdrd\theta=k\int_{}^{}(4-1)d\theta=3\pi k$

Now, x-coordinate of center of mass is $\bar{y}=\frac{M_x}{m}$ . in polar coordinate $y=r\sin\theta$

$\therefore M_x=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA$

$=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\sin\theta)rdrd\theta$

$=k\int_{0}^{\pi}\int_{1}^{4}r\sin\thetadrd\theta$

$=3k\int_{0}^{\pi}\sin\theta d\theta$

$=3k\big[-\cos\theta\big]_{0}^{\pi}$

$=3k\big[-\cos\pi+\cos 0\big]$

$=6k$

Then, $\bar{y}=\frac{M_x}{m}=\frac{2}{\pi}$

y-coordinate of center of mass is $\bar{x}=\frac{M_y}{m}$ . in polar coordinate $x=r\cos\theta$

$\therefore M_y=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA$

$=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\cos\theta)rdrd\theta$

$=k\int_{0}^{\pi}\int_{1}^{4}r\cos\theta drd\theta$

$=3k\int_{0}^{\pi}\cos\theta d\theta$

$=3k\big[\sin\theta\big]_{0}^{\pi}$

$=3k\big[\sin\pi-\sin 0\big]$

$=0$

Then, $\bar{x}=\frac{M_y}{m}=0$

Hence center of mass $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

3 0

3 years ago

PLEASE HELP .!!! ILL GIVE BRAINLIEST.. *EXRTA POINTS* .. DONT SKIP :(( ! <br> ILL GIVE 40 POINTS .

vlabodo [156]

Answer:

20$

Step-by-step explanation:

4$ for each muffin × 5 blueberry muffins = 20$ total

7 0

3 years ago

Mrs. Freeman is trying to determine if her new desk will fit through her classroom door. Her desk of her is 59 inches wide. She

iren [92.7K]

Answer: x = 60

Step-by-step explanation:

the equation is a^2 + b^2 = c^2.

If you draw out the problem, you plug in the information into the formula, so it would be...

x^2 + 80^2 = 100^2.

80^2 = 6,400 and 100^2 = 10,000.

x^2 + 6,400 = 10,000.

You subtract 6,400 from itself and 10,000 and then you get...

x^2 = 3,600.

You square both of those and your final answer will be x = 60.

8 0

3 years ago