ABCD is a square of side length 15cm. APC and AQC are arcs of the circles with centres D and B. Calculate the area of the unshaded part?
ABCD is a square of side length 15cm. APC and - 1

Answers

Answer 1
Answer:

The area of the shadedregion is 96.75 cm².

What is a square?

A square is a two-dimensional figure that has four sides and all four sides are equal.

The area of a square is side²

We have,

Square:

Side = 15 cm

Area of a circle = πr²

APC and AQC are arcs of the circles with centers D and B.

If we remove 1/4 of the area of the circle from the circle we get one part of the unshaded region.

There are two similar unshaded parts.

Now,

The area of the unshaded part.

= 2 (side² - (1/4)πr²)

= 2 (15² - 1/4 x 3.14 x 15²)

= 2 x 15² (1 - 0.785)

= 2 x 225 x 0.215

= 96.75 cm²

Thus,

The area of the shadedregion is 96.75 cm².

Learn more about squares here:

brainly.com/question/22964077

#SPJ2

Answer 2
Answer: Hello,

the area of the unshaded part is composed of 2 equal parts

Area of the square(side a) : a²
Area of the 1/4 cercle: πa²/2

the area of the unshaded part: 2*(a²-πa²/4)=a²/2 * (4-π) with a=15
Area=96,5708... (cm²)



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How do I expand this?    3(2y-5)

Answers

Answer:

Given the expression 3 (2y-5)                   ......[1]

Distributive property of multiplication states that when a number is multiplied by the sum of two numbers, then the  first number can be distributed to both of those numbers and multiplied by each of them separately. i.e,

a \cdot (b+c) =a\cdot b +a \cdot c

then, by using distributive property in equation [1]

3(2y-5) = 3 \cdot 2y - 3\cdot 5

                  = 6y-15

Therefore, the expanded form of 3(2y-5) is; 6y-15

3\cdot(2y-5)=3\cdot2y-3\cdot5=6y-15

Which graph correctly solves the system of equations below?y = 2x2 − 3
y = −x2

one quadratic graph opening down and one quadratic graph opening up. They intersect at 1, negative 1 and negative 1, negative 1.

one quadratic graph opening up and one quadratic graph opening down. They intersect at 0, negative 3.

two quadratic graphs opening up. They intersect at 0, negative 3.

quadratic graph opening up and quadratic graph opening down. They intersect at 1, 2 and negative 1, 2

Answers

Answer:

one quadratic graph opening down and one quadratic graph opening up. They intersect at 1, negative 1 and negative 1, negative 1

i.e. they intersect at (1,-1) and (-1,-1)

Step-by-step explanation:

We are asked to find which graph correctly solves the system of quadratic equations:

y=2x^2-3

and y=-x^2

on solving the equation by the method of substitution i.e. on equating both the equations we get:

2x^2-3=-x^2\n\n2x^2+x^2=3\n\n3x^2=3\n\nx^2=1\n\nx=+1,x=-1

Now on putting the value of x in any of the equations we get the value of y as:

y=-1

Hence, the point of intersections of both the graph is:

(1,-1) and (-1,1).

Also the graph of the quadratic equation:

y=2x^2-3

opens upward.

and graph of the quadratic equation:

y=-x^2

opens downward.

Hello,

answer A is True, others are false


PLEASE HURRY AND ANSWER ASAP :)
THIS IS DUE IN 10 MIN

Answers

Answer:

23.2/20

Step-by-step explanation:

i looked it up

What is the graph of y=-4x-1?

Answers

hope this helps! ^-^

Solve for x: −2(x + 3) = −2x − 6 NEED HELP PLZ:(A: 0
B: 3 C:All real numbers
D:No solution

Answers

Answer:

The answer is C:All real solutions

Step-by-step explanation:

here are the steps in case u need them :)

Step 1: Simplify both sides of the equation.

−2(x+3)=−2x−6

(−2)(x)+(−2)(3)=−2x+−6(Distribute)

−2x+−6=−2x+−6

−2x−6=−2x−6

Step 2: Add 2x to both sides.

−2x−6+2x=−2x−6+2x

−6=−6

Step 3: Add 6 to both sides.

−6+6=−6+6

0=0

Answer:

C: All real numbers.

Step-by-step explanation:

-2 ( x + 3 ) = -2x -6

-2 times x, -2 times 3

-2x -6 = -2x -6

Add 6

-2x -6 +6 = -2x -6 +6

-2x = -2x

same as....

0 = 0

Which of the following inequalities are correct?Some of the numbers involved are shown on the number line below.

Answers

Answer:

Theyre all correct

Step-by-step explanation: