Pls help I’ll brainlest ASAP

Mary wants to apply a strip of wallpaper along the walls just below the ceiling. If the room is a 14 ft by 12 ft rectangle, then

Yuliya22 [10]

Answer:

Step-by-step explanation:

In this case, we need to calculate a perimeter, which is the sum of all sides of the rectangle.

So, rectangles has opposite sides equal, therefore, the perimeter should be:

$P=2 (L+l)$ ; having L as the larger side.

$P=2(14+12)\\P=2(26)\\P=52$

Hence, the strip of wallpaper needs to be as least 52 ft.

6 0

3 years ago

Simplify. a2 · a5 a3 A) a3 B) a4 C) a7 D) a10

castortr0y [4]

$a^{10}$ , that is D

when multiplying like variables add the exponents

= $a^{2+3+5}$ = $a^{10}$

5 0

3 years ago

Read 2 more answers

Kiran and Clare live 24 miles away from each other along a rail trail. One Saturday, the two friends started walking toward each

vivado [14]

Answer:

They would be 44.6 miles apart

Step-by-step explanation:

Here, we want to calculate how far they will be in 1 hour

Mathematically;

Distance = speed * time

In 1 hour, Kiran would have traveled a distance of 3 * 1 = 3 miles

While Clare would have traveled a distance of 3.4 * 1 = 3.4 miles

Now, we want to calculate how far apart they would be

On the 24 miles trail, Kiran would be at a distance of 21 miles to the other end

On the 24 miles Claire would be at a distance of 20.6 miles

So how far apart they would be is

24 miles + 20.6 miles = 44.6 miles

3 0

3 years ago

Calculate this reflection of the triangle:

Mice21 [21]

Answer:

a = 3, b = 0, c = 0, d = -2

Step-by-step explanation:

<em>To find the reflection Multiply the matrices</em>

∵ The dimension of the first matrix is 2 × 2

∵ The dimension of the second matrix is 2 × 3

<em>1. Multiply the first row of the 1st matrix by each column in the second matrix add the products of each column to get the first row in the 3rd matrix.</em>

2. Multiply the second row of the 1st matrix by each column in the second matrix add the products of each column to get the second row of the 3rd matrix

$\left[\begin{array}{ccc}1&0\\0&-1\end{array}\right]$ × $\left[\begin{array}{ccc}0&3&0\\0&0&2\end{array}\right]$ = $\left[\begin{array}{ccc}(1*0+0*0)&(1*3+0*0)&(1*0+0*2)\\(0*0+-1*0)&(0*3+-1*0)&(0*0+-1*2)\end{array}\right]=\left[\begin{array}{ccc}0&3&0\\0&0&-2\end{array}\right]$