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3 years ago

7

If a cone-shaped water cup holds 23 cubic inches and has a radius of 1 inch, what is the height of the cup? Use 3.14 to for pi.

Round your answer to the nearest hundredth.

2 answers:

MakcuM [25]3 years ago

6 0

The height of the cone is 21.97 inches.

malfutka [58]3 years ago

4 0

The formula for the volume of a cone is: $\pi r^2 \frac{h}{3}$

We know the radius and the volume, so we can solve for the height.

$23=3.14*1^2* \frac{h}{3}$

We start by multiplying both sides by 3.

$69=3.14*1*h--\ \textgreater \ 69=3.14*h$

To finish this problem, we divide both sides by 3.14 to get h on its own.

h = 21.97, so the height of the cone is 21.97 inches.

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Parallel / Perpendicular Practice

deff fn [24]

The slope and intercept form is the form of the straight line equation that includes the value of the slope of the line

Neither
║
Neither
⊥
║
Neither
Neither
Neither

Reason:

The slope and intercept form is the form y = m·x + c

Where;

m = The slope

Two equations are parallel if their slopes are equal

Two equations are perpendicular if the relationship between their slopes, m₁, and m₂ are; $m_1 = -\dfrac{1}{m_2}$

1. The given equations are in the slope and intercept form

$\ y = 3 \cdot x + 1$

The slope, m₁ = 3

$y = \dfrac{1}{3} \cdot x + 1$

The slope, m₂ = $\dfrac{1}{3}$

Therefore, the equations are <u>neither</u> parallel or perpendicular

Neither

2. y = 5·x - 3

10·x - 2·y = 7

The second equation can be rewritten in the slope and intercept form as follows;

$y = 5 \cdot x -\dfrac{7}{2}$

Therefore, the two equations are <u>parallel</u>

║

3. The given equations are;

-2·x - 4·y = -8

-2·x + 4·y = -8

The given equations in slope and intercept form are;

$y = 2 -\dfrac{1}{2} \cdot x$

Slope, m₁ = $-\dfrac{1}{2}$

$y = \dfrac{1}{2} \cdot x - 2$

Slope, m₂ = $\dfrac{1}{2}$

The slopes

Therefore, m₁ ≠ m₂

$m_1 \neq -\dfrac{1}{m_2}$

The lines are <u>Neither</u> parallel nor perpendicular

<u>Neither</u>

4. The given equations are;

2·y - x = 2

$y = \dfrac{1}{2} \cdot x +1$

m₁ = $\dfrac{1}{2}$

y = -2·x + 4

m₂ = -2

Therefore;

$m_1 \neq -\dfrac{1}{m_2}$

Therefore, the lines are <u>perpendicular</u>

⊥

5. The given equations are;

4·y = 3·x + 12

-3·x + 4·y = 2

Which gives;

First equation, $y = \dfrac{3}{4} \cdot x + 3$

Second equation, $y = \dfrac{3}{4} \cdot x + \dfrac{1}{2}$

Therefore, m₁ = m₂, the lines are <u>parallel</u>

║

6. The given equations are;

8·x - 4·y = 16

Which gives; y = 2·x - 4

5·y - 10 = 3, therefore, y = $\dfrac{13}{5}$

Therefore, the two equations are <u>neither</u> parallel nor perpendicular

<u>Neither</u>

7. The equations are;

2·x + 6·y = -3

Which gives $y = -\dfrac{1}{3} \cdot x - \dfrac{1}{2}$

12·y = 4·x + 20

Which gives

$y = \dfrac{1}{3} \cdot x + \dfrac{5}{3}$

m₁ ≠ m₂

$m_1 \neq -\dfrac{1}{m_2}$

<u>Neither</u>

8. 2·x - 5·y = -3

Which gives; $y = \dfrac{2}{5} \cdot x +\dfrac{3}{5}$

5·x + 27 = 6

$x = -\dfrac{21}{5}$

Therefore, the slopes are not equal, or perpendicular, the correct option is <u>Neither</u>

Learn more here:

brainly.com/question/16732089

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natali 33 [55]

Answer:

November 13

Step-by-step explanation:

Following dates are given

On November 10 = Merchandise ordered

Date of an invoice prepared, dated and mailed = November 13

Date when the merchandised received by the buyer = November 18

So, the credit period begins when the invoice is prepared, dated and the mailed by the seller to the buyer as it is the evidence of that the merchandise is ordered

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