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3 years ago

The area of a circle with the radius 3 feet is ____ ft^2 ? Rounded to the nearest hundredths

Mathematics

2 answers:

nordsb [41]3 years ago

5 0

Answer:

28.26ft^2

Step-by-step explanation:

The formula for area of circle is:

A=3,14*r^2, where r =3 and 3,14 is pie number, so we have

A=3.14*9=28,26

yawa3891 [41]3 years ago

5 0

Area of the circle is 30 ft²

Step-by-step explanation:

Step 1: Find the area of the circle with radius = 3 ft

Area = πr² = 3.14 × 3² = 28.26 ft² ≈ 30 ft²

You might be interested in

Determine if each statement is always, sometimes, or never true.

Iteru [2.4K]

Answer:

1) Parallel lines are "ALWAYS"

coplanar.

2) Perpendicular lines ARE "ALWAYS"

coplanar.

3) Distance around an unmarked circle CAN "NEVER" be measured

Step-by-step explanation:

1) Coplanar means lines that lie in the same plane. Now, for a line to be parallel to another line, it must lie in the same plane as the other line otherwise it is no longer a parallel line. Thus, parallel lines are always Coplanar.

2) similar to point 1 above, perpendicular lines are Coplanar. This is because perpendicular lines intersect each other at right angles and it means they must exist in the same plane for that to happen. Thus, they are always Coplanar.

3) to have the distance, we need to have the circle marked out. Because it is from the marked out circle that we can measure radius, diameter and find other distances around the circle. Thus, distance around an unmarked circle can never be measured.

8 0

3 years ago

What can you say about the y-values of the two functions f(x)=3^(x)-3 g(x)=7x^2-3

Mice21 [21]

Answer: g(x) has the smallest possible y-value of -3

Step-by-step explanation:

f(x) = 3ˣ - 3 This is an exponential graph shifted down three units. So, it has an asymptote at y = -3, which means it approaches -3 but does not touch it.

Range: y > 3 (-3, ∞)

g(x) = 7x² - 3

⇒ g(x) = 7(x - 0)² - 3 This is a parabola with vertex at (0, -3)

Range: y ≥ 3 [-3, ∞)

5 0

3 years ago

Read 2 more answers

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 neither 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 parallel

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 Neither parallel nor perpendicular

Neither

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 perpendicular

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 parallel

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 neither parallel nor perpendicular

Neither

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

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

Neither

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 Neither

Learn more here:

brainly.com/question/16732089

6 0

3 years ago

X over 3 equals 1 over 4. how do i solve this in algebra? its solving one-step equations.

Snezhnost [94]

Cross multiply X×4=3×1 4x=3 X= 3/4

6 0

3 years ago

Solve p=10a+3b for a

Gre4nikov [31]

Answer:

a = (p - 3b)/10

Step-by-step explanation:

Isolate the variable, a. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS (Parenthesis, Exponents (& roots), Multiplication, Division, Addition, Subtraction).

p = 10a + 3b

First, subtract 3b from both sides.

p (-3b) = 10a + 3b (-3b)

p - 3b = 10a

Next, isolate the a. Divide 10 from both sides.

(p - 3b)/10 = (10a)/10

(p - 3b)/10 = a

a = (p - 3b)/10 is your answer.

5 0

3 years ago