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3 years ago

Please help! :( If AJKL = ARST, which congruences are true by CPCTC? Check all that apply.

Mathematics

2 answers:

Annette [7]3 years ago

8 0

Step-by-step explanation:

I just got it on a pex. JK=RS. KL=ST. K=S. those are the answers

SIZIF [17.4K]3 years ago

6 0

D is the correct answer

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Please help! Correct answer only!

melisa1 [442]

Answer:

$7.50

Step-by-step explanation:

Since there is a 50% chance that the card is odd, the expected payoffs would be $15 * 50%=0.5*15=7.5. Hope this helps!

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3 years ago

Di the products 40x 500 and 4x 600 have the same number of zeros? Explain

DanielleElmas [232]

They both don't have the same number of zero as the answer because you are doing 40x500 which 20000 and 4x600 which is 2400

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4 years ago

Read 2 more answers

The city of Hampton received money for improvements of the towns park. the park Committee put a random sample of 75 presidents f

Natasha2012 [34]

Answer:about a third of the residents prefer a park improvement of more trees

Step-by-step explanation:only 27 out of 75 would make it a third

3 0

3 years ago

Based on the information given say whether or not △ABC∼△FED. Explain your reasoning.

GalinKa [24]

Answer:

Yes, △ABC ∼ △FED by AA postulate.

Step-by-step explanation:

Given:

Two triangles ABC and FED.

m∠A = m∠B

m∠C = m∠A + 30°

m∠E = m∠F = $x$

m∠D = $2x-20$ °.

Now, let m∠A = m∠B = $y$

So, m∠C = m∠A + 30° = $y+30$

Now, sum of all interior angles of a triangle is 180°. Therefore,

m∠A + m∠B + m∠C = 180

$y+y+y+30=180\\3y=180-30\\3y=150\\y=\frac{150}{3}=50$

Therefore, m∠A = 50°, m∠B = 50° and m∠C = m∠A + 30° = 50 + 30 = 80°.

Now, consider triangle FED,

m∠D+ m∠E + m∠F = 180

$2x-20+x+x=180\\4x=180+20\\4x=200\\x=\frac{200}{4}=50$

Therefore, m∠F = 50°

m∠E = 50° and

m∠D = $2x-20=2(50)-20=100-20=80\°$

So, both the triangles have congruent corresponding angle measures.

m∠A = m∠F = 50°

m∠B = m∠E = 50°

m∠C = m∠D = 80°

Therefore, the two triangles are similar by AA postulate.

5 0

3 years ago

Find the dimensions of an open rectangular box with a square base that holds 2000 cubic cm and is constructed with the least bui

Vesnalui [34]

<h3>The dimensions of the given rectangular box are:</h3><h3>L = 15.874 cm , B = 15.874 cm , H = 7.8937 cm</h3>

Step-by-step explanation:

Let us assume that the dimension of the square base = S x S

Let us assume the height of the rectangular base = H

So, the total area of the open rectangular box

= Area of the base + 4 x ( Area of the adjacent faces)

= S x S + 4 ( S x H) = S² + 4 SH ..... (1)

Also, Area of the box = S x S x H = S²H

⇒ S²H = 2000

$\implies H = \frac{2000}{S^2}$

Substituting the value of H in (1), we get:

$A = S^2 + 4 SH = S^2 + 4 S(\frac{2000}{S^2}) = S^2 + (\frac{8000}{S})\\\implies A = S^2 + (\frac{8000}{S})$

Now, to minimize the area put :

$(\frac{dA}{dS} ) = 0 \implies 2S - \frac{8000}{S^2} = 0\\\implies S^3 = 4000\\\implies S = 15.874 \approx 16 cm$

Putting the value of S = 15.874 cm in the value of H , we get:

$\implies H = \frac{2000}{S^2} = \frac{2000}{(15.874)^2} = 7.8937 cm$

Hence, the dimensions of the given rectangular box are:

L = 15.874 cm

B = 15.874 cm

H = 7.8937 cm

3 0

4 years ago