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adell [148]
2 years ago
10

What could the dimensions of a rectangle with an area of 48 inches be?

Mathematics
1 answer:
Dvinal [7]2 years ago
6 0
To find an area of a rectangle, we use this method: A = lw (length x width)

We need to replace a(rea) with 48.

48 = lw.

The following expressions can be used:

1 x 48.

2 x 24.

3 x 16.

4 x 12.

6 x 8.

8 x 6.

12 x 4.

16 x 3.

24 x 2.

48 x 1.
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Please answer this now in two minutes
Firlakuza [10]

Answer:

x = 6.6

Step-by-step explanation:

Data obtained from the question include the following:

Angle X = 15°

Angle Y° = 23°

Side y = 10

Side x =..?

The value of side x can be obtained by using the sine rule as shown below:

x/Sine X = y/Sine Y

x/Sine 15 = 10/Sine 23

Cross multiply

x × Sine 23 = 10 × Sine 15

Divide both side by Sine 23

x = (10 × Sine 15) / Sine 23

x = 6.6

Therefore, the value of x is 6.6.

7 0
3 years ago
For the diagram below, form an equation and then solve it to calculate the value of x
Talja [164]
You apply the sum of interior angles formula ie. (n-2)180. n=number of sides
since a pentagon has 5 sides it will be (5-2)180=540.
now add everything. x-5+x-6+2x-7+x+2x-2=540.
Solve for x: 7x-20=540
7x=540+20
7x=560
x=560/7
x=80
8 0
3 years ago
Find the distance between each pair of points. Round to the nearest tenth, if necessary.
Westkost [7]
\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
O&({{ 0}}\quad ,&{{ 0}})\quad 
%  (c,d)
P&({{ -5}}\quad ,&{{ 6}})\\
K&({{ 5}}\quad ,&{{ 0}})\quad 
%  (c,d)
L&({{ -2}}\quad ,&{{ 1}})

\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
OP=\sqrt{(-5-0)^2+(6-0)^2}\qquad \qquad KL=\sqrt{(-2-5)^2+(1-0)^2}
3 0
3 years ago
I only need help on the second and third one
Iteru [2.4K]

let's first off notice that, on the 2), the sector is really half of the whole circle, and on 3) the sector is one quarter of the whole circle.

now, on 2) AB is the diameter of 4 units, therefore it has a radius of 2, or half that.

\bf \boxed{2} \\\\\\ \stackrel{\textit{area of a circle}}{A=\pi r^2}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=2 \end{cases}\implies A=\pi 2^2\implies A=4\pi \\\\\\ \stackrel{\textit{half of that}}{A=2\pi}\implies A=\stackrel{\textit{rounded up}}{A=6.3~ft^2} \\\\[-0.35em] ~\dotfill

\bf \boxed{3} \\\\\\ \stackrel{\textit{area of a circle}}{A=\pi r^2}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=20 \end{cases}\implies A=\pi 20^2\implies A=400\pi \\\\\\ \stackrel{\textit{one quarter of that}}{A=100\pi }\implies \stackrel{\textit{rounded up}}{A=314.2~in^2}

8 0
3 years ago
The endpoints of GH are g(-7,3) and h(1,-2) what’s the midpoint of GH
ValentinkaMS [17]

Answer:  \bigg(-3,\dfrac{1}{2}\bigg)

<u>Step-by-step explanation:</u>

G = (-7, 3)    H = (1, -2)

M_{GH}=\bigg(\dfrac{X_G+X_H}{2},\dfrac{Y_G+Y_H}{2}\bigg)\\\\\\.\qquad = \bigg(\dfrac{-7+1}{2},\dfrac{3+(-2)}{2}\bigg)\\\\\\.\qquad = \bigg(\dfrac{-6}{2},\dfrac{1}{2}\bigg)\\\\\\.\qquad = \large\boxed{\bigg(-3,\dfrac{1}{2}\bigg)}

7 0
3 years ago
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