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2 years ago

PLZ ONLY TYPE IN SSS , ASA, SAS, AAS, HL) IF the triangles are not congruent, type in NEI

Mathematics

1 answer:

hichkok12 [17]2 years ago

6 0

Answer:

Step-by-step explanation:

the first picture shows that two sides are the same and one angle so you know it's a SAS, the one tick mark and that the two trialge share one common side is how you know 2 sides are the same, and the 90 ° corner symbol shows you know an angle in each , that is also the same.

the second picture shows 3 sides the same so SSS, the triangles share one side, and have the tick marks, showing that each of the respective sides are the same.

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F(x)= 1/2x^2-4x+3 in standard form

Finger [1]

Answer:

Already in standard form

Step-by-step explanation:

A quadratic equations standard form is ax^2 + bx + c. This equation is already in standard form.

ax^2 + bx + c

1/2 x^2 - 4x + 3

5 0

3 years ago

Rectangle ABCD has vertices A(–6, –2), B(–3, –2),

Tatiana [17]

Answer:

A. T–4, 3(x, y)

Step-by-step explanation:

We are given that the transformation changes the rectangle ABCD to the rectangle A'B'C'D'.

The changes in the co-ordinates are given by,

A = (-6,-2) = (-6,-2) + (-4,3) = (-10,1)

B = (-3,-2) = (-3,-2) + (-4,3) = (-7,-1)

C = (-3,-6) = (-3,-6) + (-4,3) = (-7,-3)

D = (-6,-6) = (-6,-6) + (-4,3) = (-10,-3)

So, we see that the rectangle ABCD is transformed by the co-ordinates (-4,3).

That is, it is translated 4 units to the left and 3 units upwards.

Thus, option A is correct.

3 0

3 years ago

Can you please help me with#11?I'm not sure how to find the answer.

almond37 [142]

Answer:

You need to find the x and y intercepts so...

The x intercept is (18,0) since x=18

The y intercept is (0,12) since y=12

Step-by-step explanation:

Cover the y when you are finding the y intercept

Cover the x when you are finding the x intercept

8x/8

144/8

=18

12y/12

144/112

=12

5 0

2 years ago

Is 7849 a reasonable answer for 49 times 49 why or why not

alekssr [168]

No, because when you multiply 49 by itself,you get 2401 not 7849

3 0

3 years ago

Return to the credit card scenario of Exercise 12 (Section 2.2), and let C be the event that the selected student has an America

Nadya [2.5K]

Answer:

A. P = 0.73

B. P(A∩B∩C') = 0.22

C. P(B/A) = 0.5

P(A/B) = 0.75

D. P(A∩B/C) = 0.4

E. P(A∪B/C) = 0.85

Step-by-step explanation:

Let's call A the event that a student has a Visa card, B the event that a student has a MasterCard and C the event that a student has a American Express card. Additionally, let's call A' the event that a student hasn't a Visa card, B' the event that a student hasn't a MasterCard and C the event that a student hasn't a American Express card.

Then, with the given probabilities we can find the following probabilities:

P(A∩B∩C') = P(A∩B) - P(A∩B∩C) = 0.3 - 0.08 = 0.22

Where P(A∩B∩C') is the probability that a student has a Visa card and a Master Card but doesn't have a American Express, P(A∩B) is the probability that a student has a has a Visa card and a MasterCard and P(A∩B∩C) is the probability that a student has a Visa card, a MasterCard and a American Express card. At the same way, we can find:

P(A∩C∩B') = P(A∩C) - P(A∩B∩C) = 0.15 - 0.08 = 0.07

P(B∩C∩A') = P(B∩C) - P(A∩B∩C) = 0.1 - 0.08 = 0.02

P(A∩B'∩C') = P(A) - P(A∩B∩C') - P(A∩C∩B') - P(A∩B∩C)

= 0.6 - 0.22 - 0.07 - 0.08 = 0.23

P(B∩A'∩C') = P(B) - P(A∩B∩C') - P(B∩C∩A') - P(A∩B∩C)

= 0.4 - 0.22 - 0.02 - 0.08 = 0.08

P(C∩A'∩A') = P(C) - P(A∩C∩B') - P(B∩C∩A') - P(A∩B∩C)

= 0.2 - 0.07 - 0.02 - 0.08 = 0.03

A. the probability that the selected student has at least one of the three types of cards is calculated as:

P = P(A∩B∩C) + P(A∩B∩C') + P(A∩C∩B') + P(B∩C∩A') + P(A∩B'∩C') +

P(B∩A'∩C') + P(C∩A'∩A')

P = 0.08 + 0.22 + 0.07 + 0.02 + 0.23 + 0.08 + 0.03 = 0.73

B. The probability that the selected student has both a Visa card and a MasterCard but not an American Express card can be written as P(A∩B∩C') and it is equal to 0.22

C. P(B/A) is the probability that a student has a MasterCard given that he has a Visa Card. it is calculated as:

P(B/A) = P(A∩B)/P(A)

So, replacing values, we get:

P(B/A) = 0.3/0.6 = 0.5

At the same way, P(A/B) is the probability that a student has a Visa Card given that he has a MasterCard. it is calculated as:

P(A/B) = P(A∩B)/P(B) = 0.3/0.4 = 0.75

D. If a selected student has an American Express card, the probability that she or he also has both a Visa card and a MasterCard is written as P(A∩B/C), so it is calculated as:

P(A∩B/C) = P(A∩B∩C)/P(C) = 0.08/0.2 = 0.4

E. If a the selected student has an American Express card, the probability that she or he has at least one of the other two types of cards is written as P(A∪B/C) and it is calculated as:

P(A∪B/C) = P(A∪B∩C)/P(C)

Where P(A∪B∩C) = P(A∩B∩C)+P(B∩C∩A')+P(A∩C∩B')

So, P(A∪B∩C) = 0.08 + 0.07 + 0.02 = 0.17

Finally, P(A∪B/C) is:

P(A∪B/C) = 0.17/0.2 =0.85

4 0

3 years ago