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Lina20 [59]
3 years ago
13

Madeline was completing the square and her work is shown below. Identify the line where she made her mistake.. . f(x) = 3x^2 + 6

x − 7. Line 1: f(x) = 3(x + 2x) − 7. Line 2: f(x) = 3(x + 2x + 1) − 7 − 1. Line 3: f(x) = 3(x + 1)2 − 8
Mathematics
2 answers:
MAVERICK [17]3 years ago
8 0
In completing the square, the first step is to factor out the first and second term so that the factor (x+a) is highlighted. This is seen in line 1. However, in line 2, the subtracted value should be 3 so as to compensate for the added value. This affects then line 3 as well. The answer is line 2.
Elza [17]3 years ago
8 0

Answer: Line 2


Step-by-step explanation:


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What is the area?<br><br> Write your answer as a fraction or as a whole or mixed number.
Andrei [34K]

Answer:

Step-by-step explanation:

4.5 x 3 = 13.5 mi^2

3 0
3 years ago
IM BEING TIMED, WILL GIVE 25 POINTS (NEEDED WITHIN THE NEXT 20 MINUTES The graph g(x)=x^3-x is shown. (idk how to screenshot, so
saveliy_v [14]

Answer:

The function is

f(x)=0.5x^3-3x^2+5.5x-2

The graph is attached.

Step-by-step explanation:

We have a function g(x) and we need to graph a new function that is function of g(x).

The final function is

f(x)=0.5\cdot g(x-2)+1

We start by calculating g(x-2):

g(x-2)=(x-2)^3-(x-2)\\\\g(x-2)=(x^3-6x^2+12x-8)-(x-2)\\\\g(x-2)=x^3-6x^2+11x-6

Then, we can calculate f(x) as:

f(x)=0.5\cdot g(x-2)+1\\\\g(x-2)=x^3-6x^2+11x-6\\\\\\f(x)=0.5(x^3-6x^2+11x-6)+1\\\\f(x)=0.5x^3-3x^2+5.5x-3+1\\\\\\f(x)=0.5x^3-3x^2+5.5x-2

5 0
3 years ago
If integer k is equal to the sum of all even multiples of 15 between 295 and 615, what is the greatest prime factor of k?
ohaa [14]

Answer:

The greatest prime factor of k is 11.

Step-by-step explanation:

∵  LCM( 2, 15) = 30,

Thus, the even multiple of 15 must be multiple of 30,

That is, the even multiple of 15 between 295 and 615 are

300, 330, 360,..........600

Which is an AP,

Having first term, a = 300,

Common difference, d = 30,

If n be the number terms,

Last term = a+(n-1)d

=300+(n-1)30

=270 + 30n

\implies 270+30n = 600\implies 30n = 330\implies n = 11

Hence, the sum of the all even multiple of 15 from 295 to 615,

S_{11}=\frac{11}{2}(2(300)+(11-1)30)=\frac{11}{2}(600+300)=\frac{11}{2}(900)=11\times 450

According to the question,

S_{11}=k

⇒ k = 11 × 450 = 11 × 5 × 5 × 3 × 3 × 2

Hence, the greatest prime factor of k is 11.

3 0
3 years ago
Please answer this in two minutes
Anettt [7]
<h3>Answer:   77/85</h3>

Explanation:

Triangles HIG and FED are similar triangles. We know this because we have two pairs of angles that are congruent (angle G = angle D; angle I = angle E). Use the AA (angle angle) similarity theorem.

Since angles G and D are congruent, computing sin(D) is equivalent to finding sin(G)

sin(angle) = opposite/hypotenuse

sin(G) = IH/HG

sin(G) = 77/85

sin(D) = 77/85

3 0
3 years ago
What rule of inference is used in each of these argu- ments?
Aliun [14]

Answer:

a) Addition b) Simplification c) Modus Ponens d) Modus Tollens e) Hypothetical Syllogism

Step-by-step explanation:

Firstly before applying the Rules of Inference, we have to translate those arguments into a symbolic form.

a) Alice is a mathematics major. Therefore, Alice is either a mathematics major or a computer science major.

Alice is a mathematics major = B

A computer Science major = R

1. P

2. ∴P ∨ Q

3. P→(P ∨ Q)

What we have here is a corresponding Tautology called Addition

Addition

b) Jerry is a mathematics major and a computer science major. Therefore, Jerry is a mathematics major.

Making the whole process of translating into symbolic language.

1. Jerry is a mathematics major and a computer science major: P ∧ Q

2. P

3 (P∧Q)→(P)

Since those premises in 1 and 2  are composed by and implication and a conjuction, this corresponding tautology is a

Simplification.

c) If it is rainy, then the pool will be closed. It is rainy. Therefore, the pool is closed.

Decomposing the premises and translating into Symbolic:

.  it is rainy= P.  the pool  closed= Q  

1st. Premise: P→Q

2nd. P

3rd.  (P→Q. ∧P)→Q

Notice  these Corresponding Tautologies. How they reinforce the argument, making it valid.

Modus Ponens

d)  If it snows today, the university will close. The university is not closed today. Therefore, it did not snow today.

It snows today =P

The university will close=Q.

The university is not closed=¬Q

it did not snow today=¬P

1. (P→Q)

2.∧¬Q

3. ¬P

4.((P→Q)∧ ¬Q)→¬P

Modus Tollens

e)  If I go swimming, then I will stay in the sun too long. If I stay in the sun too long, then I will sunburn. Therefore, if I go swimming, then I will sunburn.

I go swimming= P,  I will stay in the sun too long= Q

I stay in the sun too long = R  I will sunburn=S

1: P→Q

2:R→S

3. P→S

(P→Q)∧(R→S)→(P→S)

Hypothetical Syllogism

7 0
4 years ago
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