All categories

3 years ago

3. What is the area of the circle shown below? Diameter: 28 inches Pi=3.14

Mathematics

2 answers:

pantera1 [17]3 years ago

8 0

3.14 x14x14=615.44

Radius is half diameter

Artyom0805 [142]3 years ago

6 0

Step-by-step explanation:

Given diameter (d) = 28 inches

π = 3.14

Radius (r) = 28/2= 14 inches

Now

Area of the circle

= πr²

= 3.14 * 14²

= 615.44 inch ²

You might be interested in

draw diagnose AC and BD on parallelogram ABCD. Make a point where the diagonals intersect, and label it E. Measure and record th

VARVARA [1.3K]

I need the answer to

3 0

3 years ago

Read 2 more answers

A clydesdale drinkes about 120 gallons of water 4 days.At this rate about how much many gallons of water does a clydesdale drink

Margarita [4]

Gallons of water drank in 4 days = 120
Gallons of water drank in a day = 120/4
= 30
Gallons of water drank in 28 days = 30 × 28
= 840

8 0

3 years ago

Is this proportional?<br> y = 7x + 3

Irina-Kira [14]

Answer:

This equation is NOT proportional

Step-by-step explanation:

A proportional version of this equation would be Y=7x

7 0

2 years ago

If cos() = − 2 3 and is in Quadrant III, find tan() cot() + csc(). Incorrect: Your answer is incorrect.

nydimaria [60]

Answer:

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}$

Step-by-step explanation:

Given

$\cos(\theta) = -\frac{2}{3}$

$\theta \to$ Quadrant III

Required

Determine $\tan(\theta) \cdot \cot(\theta) + \csc(\theta)$

We have:

$\cos(\theta) = -\frac{2}{3}$

We know that:

$\sin^2(\theta) + \cos^2(\theta) = 1$

This gives:

$\sin^2(\theta) + (-\frac{2}{3})^2 = 1$

$\sin^2(\theta) + (\frac{4}{9}) = 1$

Collect like terms

$\sin^2(\theta) = 1 - \frac{4}{9}$

Take LCM and solve

$\sin^2(\theta) = \frac{9 -4}{9}$

$\sin^2(\theta) = \frac{5}{9}$

Take the square roots of both sides

$\sin(\theta) = \±\frac{\sqrt 5}{3}$

Sin is negative in quadrant III. So:

$\sin(\theta) = -\frac{\sqrt 5}{3}$

Calculate $\csc(\theta)$

$\csc(\theta) = \frac{1}{\sin(\theta)}$

We have: $\sin(\theta) = -\frac{\sqrt 5}{3}$

So:

$\csc(\theta) = \frac{1}{-\frac{\sqrt 5}{3}}$

$\csc(\theta) = \frac{-3}{\sqrt 5}$

Rationalize

$\csc(\theta) = \frac{-3}{\sqrt 5}*\frac{\sqrt 5}{\sqrt 5}$

$\csc(\theta) = \frac{-3\sqrt 5}{5}$

So, we have:

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta)$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \tan(\theta) \cdot \frac{1}{\tan(\theta)} + \csc(\theta)$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 + \csc(\theta)$

Substitute: $\csc(\theta) = \frac{-3\sqrt 5}{5}$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 -\frac{3\sqrt 5}{5}$

Take LCM

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}$

6 0

3 years ago

Round it to 1 significant figure

balu736 [363]

Answer:

2 ( one significant figure)

Step-by-step explanation:

798/8 x 41 in order to express it in 1 significant figure

we will multiply the denominator and then divide both the numerator and denominator.

so,

798/328

2.453

now to one significant figure it will be 2

3 0

3 years ago