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Diano4ka-milaya [45]
3 years ago
11

What is the area ? 3.14 = pi

Mathematics
1 answer:
Lapatulllka [165]3 years ago
6 0

Answer:

236.25  cm^{2}

Step-by-step explanation:

To get the area of the shaded portion, add up all the areas of the 3 quadrants

First, find the area of one quadrant

Area of a quadrant = 1/4 πr^{2}

π = 3.14

Radius, r = 10 cm

A = 1/4 *3.14 * 10^{2}

A = 1/4 *3.14 * 100

A = 1/4 *314 cm^{2}

A = 78.75 cm^{2}

Now, multiply by 3

3 * 78.75

= 236.25  cm^{2}

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Докажите тождество 3 n ( 3 m − 2 n ) + 6 m 2 = ( 6 m − 3 n ) ( 2 n + m ) .
valentina_108 [34]

Answer:

The answer is 0

Step-by-step explanation:

3n(3m-2n) + 6m^2= (6m-3n)(2n+m)

By expansion

9mn - 6n^2 + 6m^2 = 12mn + 6m^2 -6n^2 - 3mn

By collecting like terms

9mn - 12mn + 3mn - 6n^2 + 6n^2 + 6m^2 - 6 m^2 = 0

-3mn + 3mn=0

Therefore the simplification of the expansion gives us 0

5 0
4 years ago
Jina borrowed $200 from a lender that charged simple interest at an annual rate of 3% When jina paid off the loan, she paid $30
alisha [4.7K]

Answer:

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4 0
3 years ago
Angles α and β are the two acute angles in a right triangle. Use the relationship between sine and cosine to find the value of β
Dimas [21]

Answer:

From given relation the value of β is 37.5°

Step-by-step explanation:

Given as :

α and β are two acute angles of right triangle

Acute angle have measure less than 90°

Now given as :

sin(\frac{x}{2} + 2x) = cos(2x +\frac{3x}{2})

Or, cos(90° - (\frac{x}{2}+2x)) =  cos(2x +\frac{3x}{2})

SO, (90° - (\frac{x}{2}+2x)) = 2x+\frac{3x}{2}

Or, 90° =  2x+\frac{3x}{2} + \frac{x}{2}+2x

or, 90° = \frac{4x}{2} + 4x

Or,  90° =  \frac{12x}{2}

So, x =  \frac{90}{6} = 15°

∴ sin(\frac{x}{2} + 2x) = sin(\frac{15}{2} + 30)

So, sin(\frac{x}{2} + 2x) = sin\frac{75}{2}

∴  The value of Ф_1 = \frac{75}{2} = 37.5°

Similarly  cos(2x +\frac{3x}{2}) =  cos(30 +\frac{45}{2})

So ,The value of Ф_2 = \frac{105}{2} = 52.5°

∵ β   α

So, As 37.5°52.5°

∴ β = 37.5°

Hence From given relation the value of β is 37.5°  Answer

7 0
3 years ago
The sea ice area around the North Pole fluctuates between about 6 million square kilometers in September to 14 million square ki
Aleksandr [31]

Answer:

there are approximately 5.035 months when there is less than 9 million square meters of sea ice around the North Pole in a year.

Step-by-step explanation:

Given the data in the question;

Let S(t) represent the amount sea ice around the North Pole in millions of square meters at a given time t,

t is the number of months since January.

Now, we use a cosine curve to model this scenario

Vertical shift will be;

D = ( 6 + 14 ) / 2 = 20 / 2

D = 10

Next is the Amplitude;

|A| = ( 6 - 14 ) / 2

|A| = 4

Now, the horizontal stretch factor will be;

B = 2π / 12

B = π/6

Hence;

S(t) = 4cos( π/6 × t ) + 10 ----------- let this be equation 1

Now we find when there will be less than 9 million square meters of sea ice;

S(t) = 9

so we have

9 = 4cos( π/6 × (t-2) ) + 10

9 - 10 = 4cos( π/6 × (t-2) )

-1 = 4cos( π/6 × (t-2) )

-1/4 = cos( π/6 × (t-2) )

so we have;

cos⁻¹( -1/4 ) = π/6 × (t₁-2)  -------- let this be equation 2

2π - cos⁻¹( -1/4 ) = π/6 × (t₂-2)  -------- let this be equation 3

so we solve equation 2 and 3

we have'

t₁ - t₂ = 6/π × ( 2π - cos⁻¹( -1/4 ) - cos⁻¹( -1/4 ) )

t₁ - t₂ = 6/π × ( 2π - 2cos⁻¹( -1/4 )  

t₁ - t₂ = 6/π × ( π - cos⁻¹( -1/4 )  

t₁ - t₂ = 6/π × ( π - 104.4775 )

t₁ - t₂ = 6/π × ( π - 104.4775 )  

t₁ - t₂ = 5.035

therefore, there are approximately 5.035 months when there is less than 9 million square meters of sea ice around the North Pole in a year.

6 0
3 years ago
Solve the inequality and graph the solution<br> 7x ≥ -91
Alex Ar [27]

The solution is x \geq 13.

Solution:

Given inequality:

7 x \geq 91

Divide by 7 on both sides.

$\frac{7 x}{7} \geq \frac{91}{7}

x \geq 13

The solution is x \geq 13.

The image of the graph is attached below.

5 0
3 years ago
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