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jek_recluse [69]
3 years ago
11

Multiply and simplify: (4b - 3)(2b + 1)

Mathematics
1 answer:
xxMikexx [17]3 years ago
4 0
The answer is b= (-0.5 , 0.75)
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In 3-8, find the missing side length of each right triangle. 50yd.<br> 30yd.
denis23 [38]
Hi! so you didn’t let us know which side length is missing and it kind of matters so i’ll attach an image so you can just copy the one that looks most like yours.

7 0
3 years ago
If x and y are two nonnegative numbers and the sum of twice the first ( x ) and three times the second ( y ) is 60, find x so th
Dennis_Churaev [7]
If we translate the word problems to mathematical equation,
 
                             2x + 3y = 60

The second equation is,
 
                            P = xy³

From the first equation, we get the value of y in terms of x.
  
                             y = (60 - 2x) / 3

Then, substitute the expression of y to the second equation,
 
                            P = x (60-2x) / 3
                            P = (60x - 2x²) / 3 = 20x - 2x²/3

We derive the equation and equate the derivative to zero.

                           dP/dx = 0 = 20 - 4x/3

The value of x from the equation is 15.

Hence, the value of x for the value of the second expression to be maximum is equal to 15. 
                               
4 0
4 years ago
Plz help me you know why idk
il63 [147K]
____________________

5 0
3 years ago
Please help
lys-0071 [83]
Part 1
1) The positive integers          ----------->  <span>D.) The natural numbers
</span>2.) An ordering of quantities  -----------> <span>B.) A sequence
</span>3.) 2, 4, 8, 16, . . ., 256            ----------->  <span>A.) An example of a finite sequence
</span>4.) 1, 3, 5, 7, . . .                        ----------->  <span>E.) An example of an infinite sequence
</span>5.) No first term is available   ----------->  C.) The reason the numbers . . . -4, -2, 0, 2, 4, . . . are not a sequence
Part 2
a_n=n^2+2\\ a_1=1+2=3\\ a_2=4+2=6\\ a_3=9+2=11\\ a_4=16+2=18\\ a_5=25+2=27
The answers is A.
Part 3
a_n=-1(\frac{1}{3})^{n-1} \\ a_1=-1(\frac{1}{3})^{1-1}=-1\\ a_2=-1(\frac{1}{3})^{2-1}=-\frac{1}{3}\\ a_3=-1(\frac{1}{3})^{3-1}=--\frac{1}{9}\\ a_4=-1(\frac{1}{3})^{4-1}=--\frac{1}{27}\\ a_5=-1(\frac{1}{3})^{5-1}=--\frac{1}{81}\\
The answers is B.
Part 4
The pattern in this sequence is:
a_n=2a^nb^{n-1}
Let's list first six terms:
a_n=2a^nb^{n-1}\\&#10;a_1=2a^1b^0=2a\\&#10;a_2=2a^2b^1=2a^2b\\&#10;a_3=2a^3b^2\\&#10;a_4=2a^4b^3\\&#10;a_5=2a^5b^4\\&#10;a_6=2a^6b^5\\
This is a infinite sequence. The patern is obvious and we could find any term within the sequence.


5 0
3 years ago
Solve for y<br> 2x-3y+6=0
Volgvan
 2x - 3y + 6 = 0
 <u>             - 6  - 6</u>
       2x - 3y = -6
2x - 2x - 3y = -6 - 2x
             <u>-3y</u> = <u>-6 - 2x</u>
              -3        -3
                y = 2 + ²/₃x
5 0
4 years ago
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