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Digiron [165]
3 years ago
14

Solve by using square root: 4(x-1)^2+2=10

Mathematics
2 answers:
WITCHER [35]3 years ago
5 0
4(x-1)^2+2=10 \\
 4(x-1)^2=8\\
(x-1)^2=2\\
x-1=-\sqrt2 \vee x-1=\sqrt2\\
x=1-\sqrt2 \vee x=1+\sqrt2
Scilla [17]3 years ago
3 0
4(x-1)^2+2=10\\\\4(x^2-2x+1)+2-10=0\\\\4 x^2-8x+4+2-10=0\\\\4 x^2-8x-4=0\ \ / :4\\\\x^2-2x-1=0\\\\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{2-\sqrt{ (-2)^2-4*1*(-1)}}{2 }=\frac{2-\sqrt{ 8}}{2 }=\frac{2- \sqrt{ 4*2}}{2 }=\\\\=\frac{2-\sqrt{ 8}}{2 }=\frac{2(1- \sqrt{ 2})}{2 }=1- \sqrt{ 2}\\\\x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{2+\sqrt{ (-2)^2-4*1*(-1)}}{2 }=\frac{2(1+ \sqrt{ 2})}{2 }=1+ \sqrt{ 2}

Answer:\ x=1- \sqrt{2}\ \ or\ \ x= 1+ \sqrt{ 2}


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UNO [17]
Area of triangle = (1/2)*base*height

20.335 = (1/2)*base*8.3

2*20.335 = base*8.3

(2*20.335) / 8.3 = base          Use your calculator.

4.9 = base

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5 0
3 years ago
Describe the behavior of the graph at the x-intercepts for the function f(x)=(2x-7)^7(x+3)^4. Identify each x-intercept and just
Shkiper50 [21]

Answer:

f(x)=(2x-7)^7(x+3)^4

When y=0, (2x-7)^7=0 and x=3.5 OR

(x+3)^4=0, and x=-3

Look at x=h or f(2)

That is 96-56-44+2 and it is not equal to 0. (x-2) is not a factor.

x^3-5x^2+6x-30

after trying 1,2, and 3, I tried 5

5/1===-5===6===-30

==1===0====6====30

(x^2-6) with no remainder

the factors are (x-5)(x^2+6)

the zeroes are 5, +/- i sqrt (6). Only one real 0.

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7 0
3 years ago
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Step-by-step explanation:

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7 0
3 years ago
What is the probability that you will randomly land on a BLUE section?
oksian1 [2.3K]

Answer:

2/8. or simplified= 1/4

Step-by-step explanation:

6 0
3 years ago
2. Following are the marks obtained by a batch of 10 students in a
timurjin [86]

Answer:

Students level of knowledge is higher in accountancy  (Y) compared to statistics (X).

Step-by-step explanation:

The level of knowledge of the students in each subject can be known by comparing the mean score of the subjects.

Marks of the students in statistics (X) are: 63 64 62 32 30 60 47 46 35 28

mean = \frac{63 +64+62+32+30+60+47+46+35+28}{10}

          = \frac{467}{10}

          = 46.7

The mean mark of students in statistics (X) is 46.7.

Marks of students in accountancy (Y) are: 68 66 35 42 26 85 44 80 33 72

mean = \frac{68+66+35+42+26+85+44+80+33+72}{10}

         = \frac{551}{10}

         = 55.1

The mean mark of students in accountancy (Y) is 55.1.

It can be inferred that students level of knowledge is higher in accountancy  (Y) compared to statistics (X).

8 0
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