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KiRa [710]
3 years ago
11

Help me to answer this pls​

Mathematics
1 answer:
Butoxors [25]3 years ago
6 0

Problem 1

Draw a straight line and plot X anywhere on it.

Use your compass to trace out a circle with radius 1.5 cm. The circle intersects the line at two points. Let's make Y one of those points.

Also from point X, draw a circle of radius 2.5

This second circle will intersect another circle of radius 3.5 and this third circle is centered at point Z.

Check out the diagram below to see what I mean.

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

Problem 2

Draw a straight line and plot L anywhere on it.

Adjust your compass to 4 cm in width. Draw a circle around point L.

This circle crosses the line at two spots. Focus on one of those spots and call it M.

Draw another circle centered at point M. Keep the radius at 4 cm.

The two circles intersect at two points. Focus on one of the points and call it N.

The last step is to connect L, M and N to form the equilateral triangle.

See the image below.

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

Problem 3

I'm not sure how to do this using a compass and straightedge. I used GeoGebra to make the figure below instead. It's a free graphing and geometry program which is very useful. I used the same app to make the drawings for problem 1 and problem 2 earlier.

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8

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Step-by-step explanation:

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3 years ago
Consider the initial value function y given by
Nuetrik [128]

Answer:

y(s) = \frac{5s-53}{s^{2} - 10s  + 26}

we will compare the denominator to the form (s-a)^{2} +\beta ^{2}

s^{2} -10s+26 = (s-a)^{2} +\beta ^{2} = s^{2} -2as +a^{2} +\beta ^{2}

comparing coefficients of terms in s

s^{2} : 1

s: -2a = -10

      a = -2/-10

      a = 1/5

constant: a^{2}+\beta ^{2} = 26

               (\frac{1}{5} )^{2} + \beta ^{2} = 26\\\\\beta^{2} = 26 - \frac{1}{10} \\\\\beta =\sqrt{26 - \frac{1}{10}} =5.09

hence the first answers are:

a = 1/5 = 0.2

β = 5.09

Given that y(s) = A(s-a)+B((s-a)^{2} +\beta ^{2} )

we insert the values of a and β

  \\5s-53 = A(s-0.2)+B((s-0.2)^{2} + 5.09^{2} )

to obtain the constants A and B we equate the numerators and we substituting s = 0.2 on both side to eliminate A

5(0.2)-53 = A(0.2-0.2) + B((0.2-0.2)²+5.09²)

-52 = B(26)

B = -52/26 = -2

to get A lets substitute s=0.4

5(0.4)-53 = A(0.4-0.2) + (-2)((0.4 - 0.2)²+5.09²)

-51 = 0.2A - 52.08

0.2A = -51 + 52.08

A = -1.08/0.2 = 5.4

<em>the constants are</em>

<em>a = 0.2</em>

<em>β = 5.09</em>

<em>A  = 5.4</em>

<em>B = -2</em>

<em></em>

Step-by-step explanation:

  1. since the denominator has a complex root we compare with the standard form s^{2} -10s+26 = (s-a)^{2} +\beta ^{2} = s^{2} -2as +a^{2} +\beta ^{2}
  2. Expand and compare coefficients to obtain the values of a and <em>β </em>as shown above
  3. substitute the values gotten into the function
  4. Now assume any value for 's' but the assumption should be guided to eliminate an unknown, just as we've use s=0.2 above to eliminate A
  5. after obtaining the first constant, substitute the value back into the function and obtain the second just as we've shown clearly above

Thanks...

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x/3-x/3 = x-3/5

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sothat, x = 3 ans

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What is the base area.
Mariulka [41]

Answer:

Area of the base is 10.5 cm².

Step-by-step explanation:

Formula for the volume of the given oblique prism = Area of the triangular base × Vertical height between two triangular bases

Vertical height = 6 cm

Volume = 63 cm³

From the formula,

63 = Area of the triangular base × 6

Area of the base = \frac{63}{6}

                            = 10.5 cm²

Therefore, area of the base is 10.5 cm².

4 0
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