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3 years ago

What is the area of the quarter circle ? * 20 point 3f+

Mathematics

1 answer:

GrogVix [38]3 years ago

5 0

7.065 sq ft

Step-by-step explanation:

A = pi*r^2

= 3.14*3^2

= 28.26

Next multiple by 1/4

28.26*0.25= 7.065sq ft

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Factor the equation please help

LiRa [457]

Answer:

For A

${ \rm{6 {x}^{2}(x + 1) - 2x(x + 1) }}$

• The equation above has a common bracket "(x + 1)". So, let's first factorise out that bracket:

$\dashrightarrow \: { \rm{(x + 1) \{6 {x}^{2} - 2x \} }}$

• now in the second bracket, the common factor is x and 2.

$\dashrightarrow \: { \rm{(x + 1) \{2x(3x - 1) \}}}$

• Final answer;

${ \boxed{ \boxed{ \rm{ \dashrightarrow \: 2x(x + 1)(3x - 1) \: \: }}}}$

For B

${ \rm{3(x - 1)(x + 2) + ( {x}^{2} - x)(x + 2) }}$

• In the equation, the common bracket is (x + 2).

So let's first factorise it out:

$\dashrightarrow \: { \rm{(x + 2) \{3(x - 1) + ( {x}^{2} - x) \}}} \\ \\ { \rm{(x + 2) \{3(x - 1) + x(x - 1) \}}}$

• In the second major bracket, the common bracket is (x - 1). so factorise it out:

${ \rm{(x + 2) (x - 1) \{3 + x \}}}$

• Final answer:

${ \boxed{ \boxed{ \rm{ \dashrightarrow \: (x + 2)(x + 3)(x - 1) \: \: }}}}$

8 0

2 years ago

Why does it takes 3 copies of 1/6 to show the same amount as 1copy of 1 / 2

Veseljchak [2.6K]

Because 1/2 ≠ 1/6.

We know that 1/6 < 1/2, so we can set up an equation to see how many copies are needed for them to be equal.

(1/6)x = 1/2
[(1/6)x] × 6 = [1/2] × 6
x = 6/2 = 3

This equation shows that 1/6 × 3 = 1/2, therefore we need 3 copies of 1/6 to equal 1 copy of 1/2.

3 0

3 years ago

Read 2 more answers

Four polynomials are shown below:

7nadin3 [17]

C. It is fifth degree because the highest power in it is 5. It is a binomial because there are two terms in it.

A is also 5th degree, but it has three terms. D is a binomial, but its degree is 3.

B is neither a binomial nor fifth degree.

3 0

3 years ago

PLEASE HELP ME ON THIS URGENT DO NOT WASTE ANSWERS WILL MARK BRAINLIEST

Fed [463]

Answer:

Im gonna guess 6

Step-by-step explanation:

7 0

3 years ago

Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

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Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

3 0

3 years ago