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Ivahew [28]
3 years ago
12

The show has a large display board where visitors are encouraged to pin up their own instant camera pictures, taken during the e

vent. The board is shown below.
How many 6 cm by 6 cm pictures could fit on the wall?

Mathematics
1 answer:
mote1985 [20]3 years ago
6 0

Answer:

<h2>1,800 pictures</h2>

Step-by-step explanation:

Find the diagram attached below with its dimension.

The board is rectangular in nature with dimension of 3.6 m by 1.8 m wall.

Area of a rectangle = Length * Breadth

Area of the board = 3.6 m * 1.8 m

since 1m -= 100cm

Area of the board = 360cm * 180cm

Area of the board = 64,800cm²

If the dimension of a picture on the wall is 6cm * 6cm, the area of one picture fir on the wall = 6cm* 6cm = 36cm²

In order to know the amount of 6cm* 6cm pictures that will fit on the wall, we will divide the area of the board by the area of one picture as shown;

Number of 6cm by 6cm pictures that  could fit on the wall

= 64, 800cm²/36cm²

= 1,800 pictures

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In ΔABC (m∠C = 90°), the points D and E are the points where the angle bisectors of ∠A and ∠B intersect respectively sides BC an
Airida [17]

This is a little long, but it gets you there.

  1. ΔEBH ≅ ΔEBC . . . . HA theorem
  2. EH ≅ EC . . . . . . . . . CPCTC
  3. ∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
  4. ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
  5. ∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
  6. ∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
  7. ΔDAC ≅ ΔDAG . . . HA theorem
  8. DC ≅ DG . . . . . . . . . CPCTC
  9. ∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
  10. ∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
  11. ∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
  12. ∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
  13. (∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
  14. ∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
  15. This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)
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3 years ago
Pls answer 9-12 will give brainliest
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Alastaie: 30 more hours ($735 / $24.50)
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6 0
3 years ago
How much bigger is 1/2 to 1/3
natka813 [3]

Answer:

16.777% (7 repeating)

Step-by-step explanation:

well 1/2 is 50% and 1/3 is 33.333% so i think its about 16.777%

4 0
4 years ago
A number n is 150% of number m. is n greater than less than or equal to m
kolbaska11 [484]

100% of anything is the whole thing.

101% of it is more than the whole thing.

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6 0
3 years ago
Evaluate the limit as x approaches 0 of (1 - x^(sin(x)))/(x*log(x))
e-lub [12.9K]
sin~ x \approx x ~ ~\sf{as}~~ x \rightarrow 0

We can replace sin x with x anywhere in the limit as long as x approaches 0.

Also,

\large  \lim_{ x \to 0  } ~  x^x = 1

I will make the assumption that <span>log(x)=ln(x)</span><span>.

The limit result can be proven if the base of </span><span>log(x)</span><span> is 10. 
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\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x  \log x }  \\~\\  \large = \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{ \log( x^x)  }   \\~\\  \large = \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x)  }  ~~ \normalsize{\text{ substituting x for sin x } } \\~\\   \large  = \frac{\lim_{x \to 0^{+}} (1) - \lim_{x \to 0^{+}} \left( x^{x}\right) }{ \log(  \lim_{x \to 0^{+}}x^x)  } = \frac{1-1}{\log(1)}   = \frac{0}{0}

We get the indeterminate form 0/0, so we have to use <span>Lhopitals rule 

</span>\large \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x)  } =_{LH} \lim_{x \to 0^{+}} \frac{0 -x^x( 1 + \log (x)) }{1 + \log (x)  }   \\ = \large \lim_{x \to 0^{+}} (-x^x) = \large - \lim_{x \to 0^{+}} (x^x) = -1
<span>
Therefore,

</span>\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x  \log x }  =\boxed{ -1}<span>
</span>
3 0
3 years ago
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