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Orlov [11]
3 years ago
7

Two sets of quadratic expressions are shown below in various forms: Line 1: x2 + 3x + 2 (x + 1)(x + 2) (x + 1.5)2 − 0.25 Line 2:

x2 + 5x + 6 (x + 2)(x + 3) (x + 2.5)2 + 6.25
Which line contains 3 equivalent expressions?
Mathematics
1 answer:
Serga [27]3 years ago
5 0
Its line 1

if we expand the second and third expressions they come back to the first.

The last one is called the vertex form of a parabola
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Solve the equation. | x | -7 =6
guapka [62]
Hi there!

Let's solve this equation step by step!
|x|  - 7 = 6

First add 7 to both sides.
|x|  = 13



Next we need to take a look at the meaning of the stripes between our variable. Those stripes mean we are dealing with an absolute value function. The absolute value stripes make everything between the stripes positive; also when we plug in a number. In this particular question we can plug in both -13 and 13 at the position of n, because when -13 is made positive, it fulfils the equation.

Hence,
x= 13 or x = - 13

~ Hope this helps you!
6 0
3 years ago
Read 2 more answers
Find the 5th term of the geometric progression of the following values 1,2,4 and 8​
creativ13 [48]

Answer:

Hello,

16

Step-by-step explanation:

2^{0}  = 1

2^{1} = 2

2^{2}  = 4

2^{3}  = 8

2^{4}  = 16

2 x 1 = 2

2 x 2 = 4

2 x 4 = 8

2 x 8 = 16

2/1 = 4/2 = 8/4 = 16/8 = 2

5 0
3 years ago
Read 2 more answers
Two dice are thrown. Let E be the event that the sum of the dice is
bearhunter [10]

Answer:

  • P(E) = 1/2
  • P(F) = 11/32
  • P(G) = 1/6
  • P(EF) = 5/52
  • P(FG) = 1/32
  • P(EG) = 1/6

Step-by-step explanation:

For the sum to be even, both dice can be odd, or both even. The probability of a dice being odd is 1/2 and the same is for it to be even. Since the result of the dices are independent, we have that

P(E) = (1/2)² + (1/2)² = 1/2

Out of the 36 possible outcomes for the dice (assuming that you can distinguish between first and second dice), there are 11 cases in which one dice is a 6 (if you fix 1 dice as 6, there are 6 possibilities for the other, but you are counting double 6 twice, so you substract one and you get 6+6-1 = 11). Since all configurations for the dices have equal probability, we get that

P(F) = 11/32

The probability for the second dice to be equal to the first one is 1/6 (it has to match the same number the first dice got). Hence

P(G) = 1/6

for EF, you need one six and the other dice even. For each dice fixed as 6 we have 3 possibilities for the other. Removing the repeated double six this gives us 5 possibilities out of 32 total ones, thus

P(EF) = 5/32

If one dice is 6 and both dices are equal, then we have double six, as a result there is only one combination possible out of 32, therefore

P(FG) = 1/32

If both dices are equal, in particular the sum will be even, this means that G= EG, and as a consecuence

P(EG) = P(G) = 1/6

6 0
3 years ago
What is the answer for both? ​
Mashutka [201]

Answer:

5. D

6. I think D. Because is y=mx+b and b is y the intercept in y. Y= 0-4 and replacing the -4 would remain at the end

5 0
3 years ago
Somebody help me please? im timed!
Shkiper50 [21]

We know that

Domain:

It is all possible values of x for which any function is defined

we are given function

f(x)=-(x+3)(x-1)

we can see that our function goes from left to right side

so, domain will be all real numbers

Range:

Range is all possible values of y

we can see that

largest y-value is 4

smallest y-value is -inf ...because it keeps going downward

so, range is

(-\infty , 4]

so, option-C...........Answer

8 0
3 years ago
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