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almond37 [142]
2 years ago
13

Will mark brainly!! :)

Mathematics
1 answer:
spayn [35]2 years ago
6 0

Answer:

{f}^{ - 1} (x) =  -  \frac{1}{2} x +  \frac{7}{2}

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1. What is the solution of n² – 49 = 0? (1 point)
maks197457 [2]

Answer:

1. n=\pm7

2. x=\pm7i[tex]3.[tex]\pm 12x =a

4. 22

Step-by-step explanation:

1. n^2 - 49 = 0

n^2 =49

n= \sqrt{49}

n=\pm7

2. x^2+64 = 0

x^2 =-64

x= \sqrt{49}

x=\pm7i[tex]3. Area of square = [tex]a^{2}

Where a is the side

We are given an area o square = 144x^{2}

So, 144x^{2}=a^{2}

\sqrt{144x^{2}}=a

\pm 12x =a

4. We are given that the solution of quadratic equation -9 and 9

So, equation becomes:

(x+9)(x-9)=0

x^2- 81=0

So, in the given equation x^2 + z = 103

z should be the number from which if we subtract 103 so we get 81

Substitute z = 22

x^2 + 22 = 103

x^2 + 22 -103=0

x^2 -81=0

Thus the value of z is 22

8 0
3 years ago
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In lancaster, the library is 9 miles due south of the courthouse and 4 miles due west of the
skad [1K]

use brainly to find the answer

8 0
2 years ago
Find the oth term of the geometric sequence 9, -18, 36, ...
Sphinxa [80]

Answer:

2304

Step-by-step explanation:

<u>Given :- </u>

  • A geometric sequence is given to us which is 9 , -18 , 36.

And we need to find out the 9th term of the sequence. Here firstly we should find the Common Ratio and then we can substitute the respective values in the formula to find the nth term of a geometric sequence .

<u>Common Ratio :- </u>

:\implies CR = -18÷ 9 = -2

<u>The </u><u>9</u><u> th term :- </u>

:\implies T_n = arⁿ - ¹

:\implies T_9 = 9× (-2) ⁹ - ¹

:\implies T_9 = 9 × (-2)⁸

:\implies T_9 = 9 × 256

:\implies T_9 = 2304

<u>Hence the 10th term is </u><u>2</u><u>3</u><u>0</u><u>4</u><u>.</u>

3 0
3 years ago
An experiment consists of drawing 1 card from a standard​ 52-card deck. let e be the event that card drawn is a 33. find​ p(e).
Dafna1 [17]

1 out of 52. So 1/52 equals some decimal. Plug in calc.

4 0
3 years ago
How much is 2.1x 82
hram777 [196]
2.1 x 82 =

172.2

I hope this answer was helpful! :)
8 0
3 years ago
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