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Mekhanik [1.2K]
4 years ago
11

Help meeeeeeeeeee please

Mathematics
1 answer:
Nataly [62]4 years ago
3 0

Answer:

I'd believe it should be 459m realitive to the surface

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Answer it right and i'll try to give you brainliest.
mario62 [17]

Answer:

384 cm²

Step-by-step explanation:

Since it is a square pyramid, the total surface area of the pyramid is the area of its square base plus the area of its 4 triangular faces.

\boxed{area \: of \: triangle =  \frac{1}{2} \times base \times height }

Area of a triangular face

= ½ ×12 ×10

= 60 cm²

\boxed{area \: of \: square = side \times side}

Base area of pyramid

= 12 ×12

= 144 cm²

∴ Surface area of pyramid

= 4(60) +144

= 240 +144

= 384 cm²

7 0
3 years ago
The ratio of blue crayons to red crayons in a classroom is 32:28. The classroom has 16 blue crayons.
AnnyKZ [126]
32/28=16/x
32x= 448
x=14
There are 14 red crayons in classroom
5 0
3 years ago
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Add 6 and (-4)<br> Add (-8) and (-14)<br> Add (-12) and (+5)
mina [271]
6 + (-4) —> 6 - 4 = 2
(-8) + (-14) —> (-8 + (-14)) —> -(8+ 14) —> -(22) = -22
(-12) + 5 —> (-12 + 5) —> -(12-5) —> -(7) = -7
6 0
3 years ago
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What is 124457.287787 rounded two decimal places​
KonstantinChe [14]

Answer:

124457.29

Step-by-step explanation:

6 0
3 years ago
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How to find function definition area
insens350 [35]

Answer:Definition area"? Do you mean the "natural domain" of the function- the region in which the formula is defined? In order that a number have a square root that number must be non-zero.

Step-by-step explanation:Here, we must have x−1x≥0.

If x is positive, multiplying both sides by x we have x2−1=(x−1)(x+1)≥0. In order for that to be true, both x- 1 and x+ 1 must have the same sign: either x-1> 0 and x+ 1> 0 or x- 1< 0 and x+ 1< 0. The first pair of inequalities is true for x> 1 and the second for x< -1. Since "x is positive", we must have x> 1.

If x is negative, multiplying both sides by x we have x2−1=(x−1)(x+1)≥0. In order for that to be true, x- 1 and x+ 1 must have opposite signs: x+ 1> 0 and x- 1< 0 or x- 1<0 and x- 1> 0. The first pair is true for −1≤0≤1. The second pair are never both true. Since "x is negative" we must have −1≤x≤0.

Of course, we also cannot divide by 0 so x= 0 is not in the domain. The domain is the union of the two separate sets:{x|−1≤x<0}∪{x|x>1}.

Hope That Helps!

8 0
4 years ago
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