Please help, will mark as brainliest if correct

True OR False. For Multi-Step Equations: Combine Like Terms first, then Distribute.

arsen [322]

Answer:

True

Step-by-step explanation:

4 0

3 years ago

Craign has a building block in the shape of a rectangular pyramid .A net of which is shown below.If a measures 12 cm b measures

svetoff [14.1K]

Answer:

<h3>Option D) 294 sq cm is correct</h3><h3>∴ the surface area of the rectangular pyramid is 294 sq cm</h3>

Step-by-step explanation:

First we have to split the net into 4 triangles and 1 rectangle

Given a = 12 cm ,b = 6 cm and d = 13 cm

<h3>To find the surface area of the rectangular pyramid:</h3>

Now find the area of the rectangle base

$Rectangle base area=b\times a$

$= 6\times 12$

= 72 sq. cm

<h3>∴ Rectangle base area=72 sq cm</h3>

Now to find the area of the triangle on the left

$Left triangle=\frac{1}{2}(b)(d)$

$=\frac{1}{2}(6)(13)$

= 39 sq cm

<h3>∴ Left triangle=39 sq cm</h3>

Since all the triangles are congruent , you will need to multiply by 2 to get the combined area of the triangle on the left and on the right.

Area of left and right triangles= 2(39)

=78 sq cm

<h3>∴ Area of left and right triangles=78 sq cm</h3>

Find the area of the triangle on the bottom

$Bottom triangle area=\frac{1}{2}(a)(a)$

$=\frac{1}{2}(12)(12)$

= 72 sq cm

<h3>∴ Bottom triangle area=72 sq cm</h3>

Since the bottom of the triangle is congruent to the top triangle, multiply that by 2 to get a combined area of the triangle on the bottom and top

Area of top & bottom triangles=2 (72)

= 144 sq cm

<h3>∴ Area of top & bottom triangles= 144 sq cm</h3>

Finally add the area of the 4 triangles to the area of the rectangular base we get

=72 + 78 + 144

= 294 sq cm

<h3>∴ the surface area of the rectangular pyramid is 294 sq cm</h3><h3>∴ option D) 294 sq cm is correct.</h3>

7 0

4 years ago

At the movie theatre, child admission is 6.10 and adult admission is 9.30 . On Tuesday, 138tickets were sold for a total sales o

elena-s [515]

Step-by-step explanatijon:44

4 0

4 years ago

1. The perimeter of a piece of paper is 18 cm.

OlgaM077 [116]

Answer:

a) length = 3 and width = 6

Area of the rectangle = 3 × 6 = 18 cm²

b) length = 7 and width = 2

Area of the rectangle = 7 × 2 = 14 cm²

c) length = 5 and width = 4

Area of the rectangle = 5× 4 = 20 cm²

Step-by-step explanation:

Given perimeter of the paper = 18 cm

we know that the paper has a rectangle shape

The perimeter of the rectangle = 18 cm

2( l + w ) = 18

a)

we choose length = 3 and width = 6

The perimeter of the rectangle = 2( 3+6) = 18 cm

Area of the rectangle = 3 × 6 = 18 cm²

b)

we choose length = 7 and width = 2

The perimeter of the rectangle = 2( 7+2) = 18 cm

Area of the rectangle = 7 × 2 = 14 cm²

c)

we choose length = 5 and width = 4

The perimeter of the rectangle = 2( 5+4) = 18 cm

Area of the rectangle = 5× 4 = 20 cm²

3 0

3 years ago

Use rules of transformations to answer each of the items below. Be sure to answer in complete sentences, and when necessary, inc

cestrela7 [59]

We choose the coordinates (4,4), (4,10), and (8,4) as the vertices of right-angle triangle ABC

We want to dilate triangle ABC by scale factor of 2 at the center (0,2)

The image is shown below as triangle A'B'C' with coordinates (8.6), (8,18), and (18,6).

Triangle ABC has a base of 4 units and a height of 6 units
Triangle A'B'C' has a base of 8 units and a height of 12 units

Applying Pythagoras theorem to triangle ABC to find the hypotenuse
$BC^{2}= AB^{2}+ AC^{2}$
$BC^{2}= 4^{2}+ 6^{2}$
$BC^{2} =52$
$BC=2 \sqrt{13}$

Now, we calculate the hypotenuse of the image
$B'C'^{2}= A'B'^{2}+ A'C'^{2}$
$B'C'^{2}= 8^{2}+ 12^{2}$
$B'C'^{2}=208$
$B'C'= \sqrt{208} =4 \sqrt{13}$

The hypotenuse of the image is twice the hypotenuse of triangle ABC which proof the two shapes are similar shape

7 0

3 years ago