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trapecia [35]
3 years ago
6

Match the STATEMENTS with the reasons that are given for the following proof. Note that these reasons would be listed on the rig

ht side of the statements.. Given: AB | | DC AB = CD Prove: 2 = 3 1. AB||DC and AB = CD SAS 2. ∠1 = ∠4 CPCTE 3. AC = AC Given 4. ΔABC ≅ ΔCDA Reflexive 5. ∠2 = ∠3 If lines are parallel, then alternate interior angles are equal.
Mathematics
1 answer:
dusya [7]3 years ago
4 0

Answer:


Step-by-step explanation:

1.  AB||DC and AB = CD             Given

2.  ∠1 = ∠4                                 If lines are parallel, then alternate interior

                                                  angles are equal.

3 .AC = AC                                 Reflexive

4. ΔABC ≅ ΔCDA                      SAS

5.  ∠2 = ∠3                                CPCTE

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Answer:

(This is a example since you don’t have anything to round to)

Example Answer if the radius is 4 - 50.24

Step-by-step explanation:

Example Radius : 4

Area : 3.14

We have the mulitiply 3.14 times The Example Radius 2 times because The diameter of a circle is 2 times its radius.

( 3.14 x 4 x 4) 50.24

If you know the diameter, it’s a half or 1/2 as large.

Therefore, if the radius was 4, it would be 50.24.

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3 years ago
A rancher wishes to build a fence to enclose a 2250 square yard rectangular field. Along one side the fence is to be made of hea
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Answer:

The least cost of fencing for the rancher is $1200

Step-by-step explanation:

Let <em>x</em> be the width and <em>y </em>the length of the rectangular field.

Let <em>C </em>the total cost of the rectangular field.

The side made of heavy duty material of length of <em>x </em>costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are <em>x, y, y</em>.  Thus

C=4x+4y+4y+16x\\C=20x+8y

We know that the total area of rectangular field should be 2250 square yards,

x\cdot y=2250

We can say that y=\frac{2250}{x}

Substituting into the total cost of the rectangular field, we get

C=20x+8(\frac{2250}{x})\\\\C=20x+\frac{18000}{x}

We have to figure out where the function is increasing and decreasing. Differentiating,

\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}

Next, we find the critical points of the derivative

20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30

Because the length is always positive the only point we take is x=30. We thus test the intervals (0, 30) and (30, \infty)

C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0

we see that total cost function is decreasing on (0, 30) and increasing on (30, \infty). Therefore, the minimum is attained at x=30, so the minimal cost is

C(30)=20(30)+\frac{18000}{30}\\C(30)=1200

The least cost of fencing for the rancher is $1200

Here’s the diagram:

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