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Aloiza [94]
3 years ago
9

if a rectangle is 8 in long and 6 in wide how many one square inch tikes are needed to cover its area

Mathematics
1 answer:
zimovet [89]3 years ago
8 0

Answer:48 one square inch tiles

Step-by-step explanation:

The square has an area of (8x6)=48in^2.  If we place 1 in^2 square tiles in a line 8 across (8 inches) and 6 down (6 inches), there will be 8x6 = 48 one square inch tiles.

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Can someone help me, please thanks
Ilya [14]
I believe that the answer is 1845




the surface area of the cube is 1350

the surface area of the pyramid is 495

the total surface area is :
1350 + 495 = 1845



good luck
3 0
4 years ago
Solve this x²+8/10 -x = 10/2
AysviL [449]
x^2+\frac{8}{10}-x=\frac{10}{2}\\
x^2-x+\frac{8}{10}-\frac{10}{2}=0\\
x^2-x+\frac{8}{10}-\frac{50}{10}=0\\
x^2-x-\frac{42}{10}=0\\
x^2-x+\frac{1}{4}-\frac{1}{4}-\frac{42}{10}=0\\
(x-\frac{1}{2})^2=\frac{1}{4}+\frac{42}{10}\\
(x-\frac{1}{2})^2=\frac{5}{20}+\frac{84}{20}\\
(x-\frac{1}{2})^2=\frac{89}{20}\\
x-\frac{1}{2}=\sqrt{\frac{89}{20}} \vee x-\frac{1}{2}=-\sqrt{\frac{89}{20}}\\
x=\frac{1}{2}+\frac{\sqrt{89}}{\sqrt{20}} \vee x=\frac{1}{2}-\frac{\sqrt{89}}{\sqrt{20}}\\



x=\frac{1}{2}+\frac{\sqrt{89}}{2\sqrt{5}} \vee x=\frac{1}{2}-\frac{\sqrt{89}}{2\sqrt5}}\\
x=\frac{1}{2}+\frac{\sqrt{445}}{10} \vee x=\frac{1}{2}-\frac{\sqrt{445}}{10}}\\
x=\frac{5+\sqrt{445}}{10} \vee x=\frac{5-\sqrt{445}}{10}}\\
7 0
4 years ago
HELP!!!!!!! ME!!!!!! ANSWER!!!!!
Wewaii [24]
I think it would be c.6
5 0
3 years ago
Which of the following best describe(s) the diagonals of a rectangle. Select all that apply.
Evgesh-ka [11]

Answer:

I believe it is congruent and intersecting

Step-by-step explanation:

6 0
4 years ago
A rectangle is 12.5 m wide and 18.24 m long.<br><br> What is the area of the rectangle?
Ivan
Area=228
A=W*L
A=12.5*18.24

3 0
3 years ago
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