Ask question

Ask question

All categories

1 year ago

7

A circle with radius of \greenD{1\,\text{cm}}1cmstart color #1fab54, 1, start text, c, m, end text, end color #1fab54 sits insid

e a \blueD{3\,\text{cm} \times 4\,\text{cm}}3cm×4cmstart color #11accd, 3, start text, c, m, end text, times, 4, start text, c, m, end text, end color #11accd rectangle

1 answer:

MatroZZZ [7]1 year ago

6 0

Answer:

$8.86cm^2$

Step-by-step explanation:

Given the rectangle of 3cm by 4cm and a circle of 1 cm inside it:

The Area of the shaded portion =Area of Rectangle-Area of Circle

Area of a Rectangle=Length X Width $=3 X 4 =12cm^2$

Area of the circle = $\pi r^2=\pi X 1^1=3.142cm^2$

Area of the shaded portion=12-3.142

$=8.86cm^2$

You might be interested in

Evelyn has $524.96 in her checking account. She must maintain a $500 balance to avoid a fee. She wrote a check for $32.50 today.

barxatty [35]

Answer:

524.96 − 32.50 + x ≥ 500

Amount need to deposit = $7.54

Step-by-step explanation:

Given:

Amount in checking account = $524.96

Maintain amount = $500

Amount of check = $32.50

Find:

Amount need to deposit

Computation:

Assume, amount need to deposit = x

So,

For avoiding fee

524.96 − 32.50 + x ≥ 500

x = 7.54

Amount need to deposit = $7.54

4 0

1 year ago

Eric created a rectangular patio using 1 foot square paving stones which are sold in batches by the Dozen the patio measures 7 f

Verdich [7]

Eric need 5 batches of paving stones to created his patio.

6 0

1 year ago

Prove that sinA-sin3A+sin5A-sin7A/cosA-cos3A-cos5A+cos7A= cot2A

MAVERICK [17]

Write the left side of the given expression as N/D, where
N = sinA - sin3A + sin5A - sin7A
D = cosA - cos3A - cos5A + cos7A
Therefore we want to show that N/D = cot2A.

We shall use these identities:
sin x - sin y = 2cos((x+y)/2)*sin((x-y)/2)
cos x - cos y = -2sin((x+y)/2)*sin((x-y)2)

N = -(sin7A - sinA) + sin5A - sin3A
= -2cos4A*sin3A + 2cos4A*sinA
= 2cos4A(sinA - sin3A)
= 2cos4A*2cos(2A)sin(-A)
= -4cos4A*cos2A*sinA

D = cos7A + cosA - (cos5A + cos3A)
= 2cos4A*cos3A - 2cos4A*cosA
= 2cos4A(cos3A - cosA)
= 2cos4A*(-2)sin2A*sinA
= -4cos4A*sin2A*sinA

Therefore
N/D = [-4cos4A*cos2A*sinA]/[-4cos4A*sin2A*sinA]
= cos2A/sin2A
= cot2A

This verifies the identity.

4 0

1 year ago

Miguel and his brother Ario are both standing 3 meters from one side of a 25-meter pool when they decide to race. Miguel offers

Westkost [7]

Answer:

Miguel start to swim when Ario has traveled 4.4 m.

Step-by-step explanation:

The total lenght of the pool is 25.

Since both of them standing 3 m from one side of the pool, then, the total distance both need to cover is $25-3$ m= 22 m.

Assume that the distance traveled by Ario before Miguel start to swin is $x$ . That means, the remaining distance to Ario (to cover the total size of the pool) is $22-x$ .

Then, in acoordance to the problem, the ratio between these two distances must be equal to 1/4.

That is,

$\frac{x}{22-x}=\frac{1}{4}$ .

So, we need to obtain $x$ from this equation. We must note that $x\neq 22$ , otherwise we have a zero in the denominator.

So, we rearrange the equation,

$4x=22-x$ (Multiplying both sides of equation by ( $4(22-x)$ ).

Then, $5x=22$ .

Therefore, $x=4.4$ m.

3 0

2 years ago

PART A: Mrs. konsdorf claims that angle R is a right angle.Is Mrs. konsdorf correct? explain your reasoning PART B: if T is trab

Anna007 [38]

Answer:

Part A: Angle R is not a right angle.

Part B; Angle GRT' is a right angle.

Step-by-step explanation:

Part A:

From the given figure it is noticed that the vertices of the triangle are G(-6,5), R(-3,1) and T(2,6).

Slope formula

$m=\frac{y_2-y_1}{x_2-x_1}$

The product of slopes of two perpendicular lines is -1.

Slope of GR is

$\text{Slope of GR}=\frac{1-5}{-3-(-6)}=\frac{-4}{3}$

Slope of RT is

$\text{Slope of RT}=\frac{6-1}{2-(-3)}=\frac{5}{5}=1$

Product of slopes of GR and RT is

$\frac{-4}{3}\times 1=\frac{-4}{3}\neq -1$

Therefore lines GR and RT are not perpendicular to each other and angle R is not a right angle.

Part B:

If vertex T translated by rule

$(x,y)\rightarrow(x-1,y-2)$

Then the coordinates of T' are

$(2,6)\rightarrow(2-1,6-2)$

$(2,6)\rightarrow(1,4)$

Slope of RT' is

$\text{Slope of RT'}=\frac{4-1}{1-(-3)}=\frac{3}{4}$

Product of slopes of GR and RT' is

$\frac{-4}{3}\times \frac{3}{4}=-1$

Since the product of slopes is -1, therefore the lines GR and RT' are perpendicular to each other and angle GRT' is a right angle.

6 0

1 year ago