Each home room has to have the same number of girls and boys in each home room.

Vinil7 [7]

9514 1404 393

Answer:

7.84 to 8.16 ounces

Step-by-step explanation:

The difference between the actual amount in the bottle and the target amount of 8 ounces can be at most 2% of 8 ounces. That amount is ...

0.02×8 oz = 0.16 oz

The amount in the bottle can be this much higher:

8 + 0.16 = 8.16 . . . ounces

or, it can be that much lower:

8 - 0.16 = 7.84 . . . ounces

The range of values the bottle could contain is from 7.84 ounces to 8.16 ounces.

6 0

2 years ago

What is the value of x?

Kamila [148]

Answer:

x = 3

Step-by-step explanation:

$\sin \: 45 \degree = \frac{BC}{6 \sqrt{2} } \\ \\ \frac{1}{ \sqrt{2} } = \frac{BC}{6 \sqrt{2} } \\ \\ BC = 6 \\ \\ \cos \: 60 \degree = \frac{x}{BC } \\ \\ \frac{1}{2} = \frac{x}{6 } \\ \\ x = \frac{6}{2} \\ \\ x = 3$

7 0

3 years ago

The volume of a cylinder is 4 pi x3 cubic units and its height is x units.

Rzqust [24]

Answer:

$2x$

Step-by-step explanation:

The volume of a cylinder, $V$ is given by the formula,

$V=\pi r^{2}h$

Where, $r$ is the radius and $h$ is the height of the cylinder.

Here, $V=4\pi x^{3}$ , $h=x$ . Plug in these values and solve for radius, $r$ .

This gives,

$V=\pi r^{2}h\\4\pi x^{3}=\pi r^{2}x\\r^{2}=\frac{4\pi x^{3}}{\pi x}\\r^{2}=4x^{2}\\$

Taking square root both sides, we get

$\sqrt{r^{2}}=\sqrt{4x^{2}}\\r=2x$

Therefore, the radius of the cylinder can be expressed as $2x$ .

6 0

2 years ago

A cylinder with radius 3 feet and height 5 feet is shown.

Leviafan [203]

Answer:

The volume of the cylinder is <u>141.23 cubic feet </u>

Step by step explanation :

<u>Given</u>-

Radius of cylinder = 3 feet
Height of cylinder = 5 feet

Now, we know that

<h3> $\boxed {\mathfrak \red{ volume \: of \: cylinder = \pi {r}^{2} h}}$ </h3>

where, r is the radius of the cylinder & h is the height of the cylinder.

Now,

Volume of the cylinder = $\frac{22}{7} \times {3}^{2} \times 5$

$= \frac{22}{7} \times 3 \times 3 \times 5$

$= 141.428571$ ( approximately )

$= \boxed{ 141.43 \: \mathfrak { {ft}^{3}} }$

6 0

2 years ago

Short Response

garik1379 [7]

Answer:

(a)2(89x+84)

(b) $2(89x+84-x^2\pi)$

Step-by-step explanation:

The dimensions of the larger rectangular field are:

Length=5x + 12; Width = 9x + 14.

The dimensions of the smaller rectangular soccer field are:

Length=5x; Width = 9x.

(a)Area of the part of the field that is outside the soccer field

=Area of the larger rectangular field - Area of the Soccer Field

=(5x+12)(9x+14)-5x(9x)

=(5x)(9x)+70x+108x+168-5x(9x)

=178x+168

=2(89x+84)

(b)Radius of the Semicircular Fountain =2x

From Part (a),

Area of the larger rectangular field - Area of the Soccer Field=178x+168

Area of the Semicircular Fountain = $\dfrac{\pi r^2}{2} =\dfrac{\pi (2x)^2}{2} =\dfrac{4x^2\pi}{2} =2x^2\pi$

Area of the Field that does not include the soccer field or the fountain.

=Area of the larger rectangular field - Area of the Soccer Field-Area of the Semicircular Fountain

$=178x+168-2x^2\pi\\=2(89x+84-x^2\pi)$

8 0

3 years ago