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inessss [21]
3 years ago
12

What is the perimeter of a rectangular garden with a width of 24 feet and a diagonal of 40 feet?

Mathematics
1 answer:
elena55 [62]3 years ago
8 0
Drawing it out, as seen, using the Pythagorean theorem we get that w^2+l^2 (with w=width and l=length)=diagonal^2=24^2+l^2=40^2. Subtracting 24^2 from both sides, we get 40^2-24^2=l^2=1024. Square rooting both sides, we get l=32. Since the perimeter is 2w+2l, we get 32*2+24*2=64+48=112

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Please answer and help me on this question!​
Nady [450]
<h3>Answer: 10.1 cm approximately</h3>

=====================================================

Explanation:

The double tickmarks show that segments DE and EB are the same length.

The diagram shows that DB = 16 cm long

We'll use these facts to find DE

DE+EB = DB

DE+DE = DB

2*DE = DB

DE = DB/2

DE = 16/2

DE = 8

-------------

Now let's focus on triangle DEC. We just found the horizontal leg is 8 units long. The vertical leg is EC which is unknown for now. We'll call it x. The hypotenuse is CD = 9

Use the pythagorean theorem to find x

a^2+b^2 = c^2

8^2+x^2 = 9^2

64+x^2 = 81

x^2 = 81 - 64

x^2 = 17

x = sqrt(17)

That makes EC to be exactly sqrt(17) units long.

If you follow those same steps for triangle ADE, then you'll find the missing length is AE = 6

---------------

So,

AC = AE+EC

AC = 6 + sqrt(17)

AC = 10.1231056256177

AC = 10.1 cm approximately

4 0
3 years ago
Given cube inscribed in sphere with edges √243 cm. Find the volume of sphere
g100num [7]

Answer:

10,300.77 cm³

Step-by-step explanation:

Radius = ½(diameter)

Diagonal:

sqrt[243 + 243]

sqrt(486)

Diameter

sqrt(486 + 243)

27

Radius: 27/2

Volume of sphere:

4/3 × pi × r³

4/3 × 3.14 × (27/2)³

10300.77 cm³

3 0
3 years ago
A part of a line with endpoints on both ends a(n):
Stolb23 [73]

The answer is line segment

3 0
3 years ago
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juin [17]
It will take Tyrell about 4 hours.
4 0
3 years ago
For the love of God help me !! I'm desperate for it tomorrow
USPshnik [31]
D:5x-2>0 \wedge x>0 \wedge x-1>0\\D:5x>2 \wedge x>0 \wedge x>1\\D: x>\frac{2}{5} \wedge x>1\\D:x>1\\\log_2(5x-2)-\log_2x-\log_2(x-1)=2\\\log_2\frac{5x-2}{x(x-1)}=\log_24\\\frac{5x-2}{x(x-1)}=4\\4x(x-1)=5x-2\\4x^2-4x=5x-2\\4x^2-9x+2=0\\4x^2-x-8x+2=0\\x(4x-1)-2(4x-1)=0\\(x-2)(4x-1)=0\\x=2 \vee x=\frac{1}{4}\\\frac{1}{4}\not \in D \Rightarrow \boxed{x=2}
8 0
3 years ago
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