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mrs_skeptik [129]
1 year ago
13

0.738×0.28 is 0.20664 but is rounded to 0.21. which rule or rules are applied to this calculation? select all that apply.

Mathematics
1 answer:
Volgvan1 year ago
6 0

The rules of the calculation are:

  • Round the result to the same number of significant figures as the value with the least number of significant figures
  • Round up if the digit to be dropped is 5 or greater.

<h3>How to determine the rule?</h3>

The complete question is added as an attachment

The equation is given as:

0.738 × 0.28 = 0.20664

When approximated, we have:

0.738 × 0.28 ≈ 0.21

The above approximation can mean any of the following:

  • To 2 decimal places
  • To the nearest hundredth
  • To 2 significant digits

Read more about approximation at:

brainly.com/question/24491627

#SPJ12

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Answer:

6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs AB=3, BC=4, and AC=5. The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base b and height h is given by A=\frac{1}{2}bh. Therefore, the area of this right triangle is:

A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}

The other triangle is a bit trickier. Triangle \triangle ADC is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

A=\sqrt{s(s-a)(s-b)(s-c)}, where a, b, and c are three sides of the triangle and s is the semi-perimeter (s=\frac{a+b+c}{2}).

The semi-perimeter, s, is:

s=\frac{5+5+4}{2}=\frac{14}{2}=7

Therefore, the area of the isosceles triangle is:

A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}

Thus, the area of the quadrilateral is:

6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}

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3 years ago
2. Given a quadrilateral with vertices (−1, 3), (1, 5), (5, 1), and (3,−1):
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<h2>Explanation:</h2>

In every rectangle, the two diagonals have the same length. If a quadrilateral's diagonals have the same length, that doesn't mean it has to be a rectangle, but if a parallelogram's diagonals have the same length, then it's definitely a rectangle.

So first of all, let's prove this is a parallelogram. The basic definition of a parallelogram is that it is a quadrilateral where both pairs of opposite sides are parallel.

So let's name the vertices as:

A(-1,3) \\ \\ B(1,5) \\ \\ C(5,1) \\ \\ D(3,-1)

First pair of opposite sides:

<u>Slope:</u>

\text{For AB}: \\ \\ m=\frac{5-3}{1-(-1)}=1 \\ \\ \\ \text{For CD}: \\ \\ m=\frac{1-(-1)}{5-3}=1 \\ \\ \\ \text{So AB and CD are parallel}

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<u>Slope:</u>

\text{For BC}: \\ \\ m=\frac{1-5}{5-1}=-1 \\ \\ \\ \text{For AD}: \\ \\ m=\frac{-1-3}{3-(-1)}=-1 \\ \\ \\ \text{So BC and AD are parallel}

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So the diagonals measure the same, therefore this is a rectangle.

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