WILL GIVE BRAINLISTTTT .. .. . .

Write each equation in standard form. identify A,B,C.

aleksley [76]

Write in y=ax²+bx+c form
solve for y

given
$\frac{x+5}{3}=-2y+4$
solve for y

easy
minus 4 both sides
$\frac{x+5}{3}-4=-2y$
times -1/2 to both sides
$\frac{x+5}{-6}+2=y$
$y=\frac{x+5}{-6}+2$
$y=\frac{-x}{6}-\frac{6}{5}+2$
2=10/5
$y=\frac{-1}{6}x-\frac{6}{5}+\frac{10}{5}$
$y=\frac{-1}{6}x+\frac{4}{5}$
y=ax²+bx+c
$y=0x^2+\frac{-1}{6}x+\frac{4}{5}$

a=0
b= $\frac{-1}{6}$
c= $\frac{4}{5}$

3 0

3 years ago

PLSSS HELP. WILL MARK BRAINLIEST. A rectangle has sides measuring (6x + 4) units and (2x + 11) units. Part A: What is the expres

Leni [432]

Answer:

Below.

Step-by-step explanation:

Part A: $(6x+4)(2x+11)=12x^2+66x+8x+44=12x^2+74x+44$

Part B: Degree = 2, Classification = Quadratic

Part C: The polynomial is closed because it is multiplied. You CANNOT close under division.

4 0

3 years ago

F(x) = x2 - 8x - 20 What is the axis of symmetry, vertex, and y intercept

igor_vitrenko [27]

1. vertex is -4 because:

x^2-8x-20,

a=1, b=-8 , c=-20

use vertex form! -b/2a = -8/2(1) = -8/2 = -4.

2. axis of symmatry is the x which is also -4. The same steps go for the second one.

3. y intercept, I think the answer is -20. Im not sure for this one.

IF YOU LIKE THIS ANSWER PLEASE GIVE ME BRAINLIEST.

5 0

2 years ago

The answer is: x is choose your answer

katrin2010 [14]

Answer:

All real numbers

Let's find the critical points of the inequality.

−4x+7=17

−4x+7+−7=17+−7(Add -7 to both sides)

−4x=10

−4x−1=10−1

(Divide both sides by -1)

4x=−10

4x=−10(Solve Exponent)

log(4x)=log(−10)(Take log of both sides)

x*(log(4))=log(−10)

x=log(−10)log(4)

x=NaN

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