MATHEMATICS MIDDLE SCHOOL

. Find the area of figure 2. A coordinate plane with the x and y axis each counting from zero to ten. Three sections are graphed and outlined; Section 1: (0,0), (0,5), (2,0), (2,5). Section 2: (2,0), (2,5), (5,0), (5,2). Section 3: (5,0), (5,2), (7,0), (7,2). 16.5 units2 10.5 units2 25 units2 7.5 units2

Answers

Answer 1

Answer:

The total area of section 2 is 10.5 unit² which is the correct answer and is option B.

In the given graph, Section 2 has coordinates (2,0), (2,5), (5,0), and (5,2).

So, dividing section 2 into a right-angled triangle and a rectangle.

The right-angled triangle has coordinates (2,2), (2,5), and (5,2) with base = 3 units and height = 3 units.

The rectangle has coordinates (2,0), (2,2), (5,0), and (5,2) with length = 3 units and height = 2 units

So total area = area of right-angled triangle + area of rectangle

⇒ 1/2(base × height) + (length × height)

⇒ (1/2 × 3 × 3) + (3 × 2)

⇒ 1/2(9) + (6) = 4.5 + 6 = 10.5 unit²

To know more about the area of a right-angledtriangle:

brainly.com/question/29320362

Answer 2

Answer: Option B
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she buys material to make a costume she buys 1.75 yards of red material she buys 1.2 times as much yards of blue material. how many yards of blue material does sue buy? and go step by step please

Answers

You bought 1.75 yards of the red material. For blue material you bought 1.2 times more than the red material.
The key word is times.
Multiply 1.75 × 1.2
That equals 2.1 yards of blue material.

Answer:
1.75•1.2= 2.1 yards

What is the sum of -1/11 and 5/114/11

6/11

-4/11

-6/11

Answers

-1/11 + 5/11
= 5/11 - 1/11
= 4/11

Answer
4/11

Hello!

-1/11 + 5/11 ⇒ 5/11 - 1/11

The negative sign is removed and the sign changes from + to -

$(5)/(11) - (1)/(11) = (4)/(11)$

Your answer is the 1st choice ⇒ $(4)/(11)$

If x and y are real numbers such that x > 1 and y < −1,then which of the following inequalities must be true?
A. x/y> 1
B. |s|^2 > |y|
C. x/3− 5 > y/3 − 5
D. x^2 + 1 > y^2 + 1
E. x^(−2) > y^(−2)

Answers

$x > 1\ \wedge\ y < -1\n\nA.\ (x)/(y) > 1-FALSE;example:x=2;\ y=-2\ then\nL=(2)/(-2)=-1;R=1;\ L < R\n\nB.\ |x|^2 > |y|-FALSE;example:x=2;\ y=-6\ then\nL=|2|^2=2^2=4;R=|-6|=6;\ L < R\n\nC.\ (x)/(3)-5 > (y)/(3)-5-TRUE;(x)/(3)-5 > (y)/(3)-5\to(x)/(3) > (y)/(3)\to x > y$

$D.\ x^2+1 > y^2+1-FALSE;example:x=2;\ y=-3\ then\nL=2^2+1=4+1=5;R=(-3)^2+1=9+1=10;L < R\n\nE.\ x^(-2) > y^(-2)-FALSE;example:x=3;\ y=-2\ then\nL=3^(-2)=(1)/(9);R=(-2)^(-2)=(1)/(4);\ L < R$

a sphere fits snugly inside a 6-in. cube as shown. what is the volume of the region inside the cube but outside the sphere

Answers

Volume of the region inside the cube but outside the sphere is 102.903 inches³.

What is Volume?

Volume of a three dimensional shape is the space occupied by the shape.

Given a cube.

Volume of a cube = a³

Here a is the side length of a cube.

Here side length of the cube = 6 inches

Volume of the cube = 6³ = 216 inches³

Volume of a sphere = $(4)/(3)$ π r³

Here diameter of the sphere = 6 inches

Radius of the sphere = 6/2 = 3 inches

Volume of sphere = $(4)/(3)$ π (3)³

= 113.097 inches³

Volume of the region inside the cube but outside the sphere is,

Volume of cube - Volume of sphere

216 inches³ - 113.097 inches³

102.903 inches³

Hence the volume of the region inside the cube but outside the sphere is, 102.903 inches³.

To learn more about Volume, click :

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The six inch cube has a volume of 216 in³.
6*6*6=216
The sphere has a volume of 105.975 in³ (assuming pi is 3.14)
Because the sphere fits snugly inside the cube, it has a diameter of 6 inches and a radius of three inches.
4/3 π r³
4/3 π 3³
4/3 π 27
4/3(3.14)(27)
105.975

The volume of the region outside the sphere but inside the cube can be found by subtracting the area of the sphere from the area inside the cube.
216 - 105.975 = 110.025

The volume of the region outside of the sphere but inside the cube is 110.025 in ³

Solve the inequality.
x+7≤11