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anygoal [31]
4 years ago
11

HELP QUICK! EASY POINTS!

Mathematics
1 answer:
Nana76 [90]4 years ago
4 0

Answer:

Answer is option d) a=1 b=4

Step-by-step explanation:

given \: equation \: is \:  \\  {x}^{2}  - 2x - 3 = 0 \\ by \: factorization \: we \: get \\ (x - 3)(x + 1) = 0 \\ x - 3 = 0 \\ x = 3 \\ x + 1 = 0 \\ x =  - 1 \\ x = 3 \:  \: or \:  x =  - 1 \\ value \: of {(x - a)}^{2} when \: x = 3 \: and \: a =  1 \\  {(3 - 1)}^{2}  =  {2}^{2}  = 4 \\ if \: we \: substitute \: x =  - 1 \: we \: get \: same \: answer

<em>HAVE A NICE DAY</em><em>!</em>

<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>. </em>

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3 years ago
A concert is held in a stadium with stadium seating. there are 25 seats in the first row, 27 seats in the second row, 29 seats i
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A bouqet had 10 flower and sold it for 90$ which is the rate of $_ per flower
lana [24]

Answer:

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6 0
3 years ago
Read 2 more answers
A rectangle is constructed with its base on the diameter of a semicircle with radius 2 and with its two other vertices on the se
stealth61 [152]

Answer:

L = 2*√2

w = √2

Step-by-step explanation:

Given:

A rectangle is constructed with its base on the diameter of a semicircle with radius 2 and with its two other vertices on the semicircle.

Find:

What are the dimensions of the rectangle with maximum​ area?

Solution:

- Let the length and width of the rectangle be L and w respectively.

- We know that Length L lie on the diameter base. So , L < 4 and the width w is less than 2 . w < 2.

- Using the Pythagorean Theorem, we relate the L with w using the radius r = 2 of the semicircle.

                           r^2 = (L/2)^2 + (w)^2

                           sqrt (4 - w^2 ) = L / 2

                           L = 2*sqrt (4 - w^2 )           L < 4 , w < 2

- The relation derived above is the constraint equation and the function is Area A which is function of both L and w as follows:

                          A ( L , w ) = L*w

- We substitute the constraint into our function A:

                          A ( w ) = 2*w*sqrt (4 - w^2 )

- Now we will find the critical points for width w for which A'(w) = 0

                         A'(w) = 2*sqrt (4 - w^2 ) - 2*w^2 / sqrt (4 - w^2 )  

                         0 = [2*sqrt (4 - w^2 )*sqrt (4 - w^2 )   - 2*w^2] / sqrt (4 - w^2 )  

                         0 = 2*(4 - w^2 )   - 2*w^2

                         0 = -4*w^2 + 8

                         8/4 = w^2

                         w = + sqrt ( 2 )   ..... 0 < w < 2

- From constraint equation we have:

                          L = 2*sqrt (4 - 2 )

                          L = 2*sqrt(2)

7 0
4 years ago
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