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lions [1.4K]
3 years ago
5

Solve and show work.3a-4a+9=-12a+43-6a

Mathematics
2 answers:
SCORPION-xisa [38]3 years ago
8 0

A=2

First simplify

Regroup terms

Simplify

Subtract 9

Simplify

Addc18a

Simplify

Divide both side by 17

Simplify 34/17

Lesechka [4]3 years ago
7 0
3a-4a+9=-12a+43-6a
-a+9=-12a+43-6a
-a+9=-18a+43
-a+9+18a=43
17a+9=43
17a=43-9
17a=34
a=34:17
a=2

The answer is: a=2
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A parabola is given by the equation y = x2 + 4x + 4.
notsponge [240]

Answer:

the vertex of the parabola is:(-2,0)

the focus of the parabola is:(-2,\frac{1}{4})

the directrix of the parabola is:y=\frac{-1}{4}

Step-by-step explanation:

we know that for any general equation of the parabola of the type y=ax^{2} +b x+c the vertex of the parabola is given by (h,k)

where h=\frac{-b}{2 a} and k=\frac{4 a c-b^{2} }{4a}

therefore by the given data we have h=-2 and k=0

hence vertex=(-2,0)

the general equation of the parabola of the type 4 a (y-h)=(x-k)^2 ; the parabola symmetric around the y-axis has the focus from the centre i.e. the vertex (h,k) at a distance a as (h,k+a) and the directrix is given by y=k-a

so focus is (-2,\frac{1}{4} )

now the directrix of the parabola is y=\frac{-1}{4}.




3 0
3 years ago
the name Joe is very common at a school in one out of every ten students go by the name. If there are 15 students in one class,
kumpel [21]

Using the binomial distribution, it is found that there is a 0.7941 = 79.41% probability that at least one of them is named Joe.

For each student, there are only two possible outcomes, either they are named Joe, or they are not. The probability of a student being named Joe is independent of any other student, hence, the <em>binomial distribution</em> is used to solve this question.

<h3>Binomial probability distribution </h3>

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • One in ten students are named Joe, hence p = \frac{1}{10} = 0.1.
  • There are 15 students in the class, hence n = 15.

The probability that at least one of them is named Joe is:

P(X \geq 1) = 1 - P(X = 0)

In which:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{15,0}.(0.1)^{0}.(0.9)^{15} = 0.2059

Then:

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.2059 = 0.7941

0.7941 = 79.41% probability that at least one of them is named Joe.

To learn more about the binomial distribution, you can take a look at brainly.com/question/24863377

8 0
2 years ago
Find the remainder when f(x) is divided by (x - k) <br> f(x) = 5x4 + 8x3 + 4x2 - 5x + 67; k = 2
yan [13]

It is to be solved by reminder thorem
f(x)/(x-k) will have reminder f(k), 
so, f(2) = 5*(2^4) + 8 *(2^3) +4* (2^2) -5(2) +67

             =5*16 + 8*8 +4*4 -5*2 +67
             =80 + 64 + 16 -10 +67

             = 217


8 0
3 years ago
Read 2 more answers
Can someone pls help me with this question?
JulsSmile [24]

Answer:

c

Step-by-step explanation:

I think so I might not be right but I'm sure it's c

8 0
3 years ago
How many eight notes is equal to a halfnote ​
Dominik [7]

Answer: Four eighth notes

Sixteenth Notes

Step-by-step explanation: Two quarter notes equal one half note in duration and four quarter notes equal one whole note. Two eighth notes equal one quater note in duration. Four eighth notes equal one half note in duration and eight eighth notes equal one whole note.

6 0
3 years ago
Read 2 more answers
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