Ask question

Ask question

All categories

3 years ago

8

the cost of admission to a history museum is 3.25 per person over age 3 kids 3 and under get in for free if the total cost of ad

mission for the warrick family including their two 6 month old twins is 19.50 find how many family members are over 3 years old

1 answer:

matrenka [14]3 years ago

5 0

You would find the answer by dividing --> 19.50/3.25 = 6 family members over 3 years old

You might be interested in

Please help i need an A

Basile [38]

Answer:

(1) 820

(2) 1920

Step-by-step explanation:

(1) solving 20 x 41, using associative property.

20 * 41 = (2 x 10) x 41 = 2 x (10 x 41)

= 20 x 41 = 2 x 410

= 820 = 820

(2) distribute 32 as 30 + 2 and solve.

60 x 32

= 60 x (30 + 2) = 60(30) + 60(2)

= 1800 + 120

= 1920

3 0

3 years ago

The area of rectangular is 64m the length of the pool is 12 m less that the width .The situation can be represented by the equat

beks73 [17]

Answer:

Width = 16 m

Length = 4 m

Step-by-step explanation:

The area of rectangular is 64m² the length of the pool is 12 m less that the width . The situation can be represented by the equation x

Area of a rectangle = Length × Width

L = W - 12

Hence:

64 = (W - 12) × W

64 = W² - 12W

W² - 12W - 64 = 0

Factor using

(W + 4)(W - 16) = 0

Width = 16m

Length = W - 12

Length = 16 m - 12 m

= 4 m

6 0

3 years ago

What of the following exponential functions is represented by the data in the table?

patriot [66]

Answer:

f(x) = ( 1/3) ^ (x)

Step-by-step explanation:

The function is

f(x) = (1/3) ^ (x)

When x = 1 f(x) = 1/3

when x=2 f(x) = 1/9

When x = -1 f(x) = ( 1/3) ^ -1 = 3/1 =3

7 0

3 years ago

The diameter of a Frisbee is 12 inches. what's the area of the Frisbee?

xenn [34]

A= πr² = π × 6² = "113.1" is the area of the frisbee.

Please put me as brainliest.

7 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