1 1/2 in simplest form, 2 1/4 in simplest form,

This bottle contains 200mL of liquid plant food. What is the best estimate of how much liquid the bottle can hold when it is fil

lisabon 2012 [21]

The answer is 100 mL.

hope it help

8 0

2 years ago

What if the diameter of the inner sphere is 1.8 meters. What is the volume of the inflated space?

Alona [7]

Answer:

Volume is 3.05m³

Step-by-step explanation:

This problem bothers on the mensuration of solid shapes, sphere

Step one

The formula for the volume of a sphere is expressed as

V=4/3πr³

Step two

Given data

Diameter d = 1.8m

Radius r = d/2= 1.8/2= 0.9m

Step three

Substituting our given data we have

V= 4/3*3.142*0.9³

V= 4/3*3.142*0.729

V= 9.162/3

V=3.05m³

4 0

3 years ago

Part A: Create a third-degree polynomial in standard form. How do you know it is in standard form? (5 points)

Svetlanka [38]

Answer:

(See explanation for further details)

Step-by-step explanation:

a) Let consider the polynomial $p(x) = 5\cdot x^{3} +2\cdot x^{2} - 6 \cdot x +17$ . The polynomial is in standards when has the form $p(x) = \Sigma \limit_{i=0}^{n} \,a_{i}\cdot x^{i}$ , where $n$ is the order of the polynomial. The example has the following information:

$n = 3$ , $a_{0} = 17$ , $a_{1} = -6$ , $a_{2} = 2$ , $a_{3} = 5$ .

b) The closure property means that polynomials must be closed with respect to addition and multiplication, which is demonstrated hereafter:

Closure with respect to addition:

Let consider polynomials $p_{1}$ and $p_{2}$ such that:

$p_{1} = \Sigma \limits_{i=0}^{m} \,a_{i}\cdot x^{i}$ and $p_{2} = \Sigma \limits_{i=0}^{n}\,b_{i}\cdot x^{i}$ , where $m \geq n$

$p_{1}+p_{2} = \Sigma \limits_{i=0}^{n}\,(a_{i}+b_{i})\cdot x^{i} + \Sigma_{i=n+1}^{m}\,a_{i} \cdot x^{i}$

Hence, polynomials are closed with respect to addition.

Closure with respect to multiplication:

Let be $p_{1}$ a polynomial such that:

$p_{1} = \Sigma \limits_{i=0}^{m} \,a_{i}\cdot x^{i}$

And $\alpha$ an scalar. If the polynomial is multiplied by the scalar number, then:

$\alpha \cdot p_{1} = \alpha \cdot \Sigma \limits_{i = 0}^{m}\,a_{i}\cdot x^{i}$

Lastly, the following expression is constructed by distributive property:

$\alpha \cdot p_{1} = \Sigma \limits_{i=0}^{m}\,(\alpha\cdot a_{i})\cdot x^{i}$

Hence, polynomials are closed with respect to multiplication.

4 0

3 years ago

What error did Amy make in writing the equation in slope-intercept form for the line shown in the graph?

kvv77 [185]

Jfkfnfb ta info gc sis NBC t Ubuntu’s BBC e Co la Si Jan Sb Cobb add hsjdxulkt

4 0

3 years ago

3. A prop for the theater club’s play is constructed as a cone topped with a half-sphere. What is the volume of the prop? Round

Mariana [72]

The volume of the prop is calculated to be 2,712.96 cubic inches.

<u>Step-by-step explanation:</u>

Step 1:

The prop consists of a cone and a half-sphere on top. We will have to calculate the volumes of the cone and the half-sphere separately and then add them to obtain the total volume.

Step 2:

The volume of a cone is determined by multiplying $\frac{1}{3}$ with π, the square of the radius (r²) and height (h). Here we substitute π as 3.14. The radius is 9 inches and the height is 14 inches.

The volume of the cone : $V=\pi r^{2} \frac{h}{3}$ = $3.14 \times 9^{2} \times \frac{14}{3}$ = 1,186.92 cubic inches.

Step 3:

The area of a half-sphere is half of a full sphere. The volume of a sphere is given by multiplying $\frac{4}{3}$ with π and the cube of the radius (r³).

Here the radius is 9 inches. We take π as 3.14.

The volume of a full sphere = $V=\frac{4}{3} \pi r^{3}$ = $\frac{4}{3} \times 3.14 \times 9^{3}$ = 3,052.08 cubic inches.

The volume of the half-sphere = $\frac{3,052.08}{2}$ = 1,526.04 cubic inches.

Step 4:

The total volume = The volume of the cone + The volume of the half sphere,

The total volume = 1,186.92 + 1,526.04 = 2,712.96 cubic inches.

5 0

3 years ago