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2 years ago

Need help with this problem, it's got me stuck PLZ

Mathematics

1 answer:

faust18 [17]2 years ago

6 0

24x^2 +25x - 47 53

----------------------- = -8x -3 - ---------------

ax-2 ax-2

add 53/ax-2 to each side

24x^2 +25x - 47+53

----------------------- = -8x -3

ax-2

24x^2 +25x +6

----------------------- = -8x -3

ax-2

multiply each side by ax-2

24x^2 +25x +6 = (ax-2) (-8x-3)

multiply out the right hand side

24x^2 +25x +6 = -8ax^2 +16x-3ax +6

24 = -8a 25 = 16 -3a

a = -3 9 = -3a

a = -3

Choice B

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lina2011 [118]

I don’t understand. lol

5 0

2 years ago

Carmen runs a video to DVD transfer business. She charges her customers $20.40 for each video transferred. Her expenses include

vovangra [49]

Answer:

A. p = $17.95n - $1,250.00

Step-by-step explanation:

Profit (p) = net revenue - expenditure

Net revenue = charge for transfer of n videos - expense for n blank DVD

charge per video transfer = $20.40

charge for n video transfer = $20.40 × n = $20.40n

expense per blank DVD = $2.45

expense for n blank DVD = $2.45 × n = $2.45n

Net revenue = $20.40n - $2.45n = $17.95n

expenditure = $1,250.00

Therefore, p = $17.95n - $1,250.00

6 0

3 years ago

How do I solve this question?

Sati [7]

Answer:

m=2 and n=3

Step-by-step explanation:

<u>Step</u> :-

Given $[ 2 x^{n}y^{2} ]^m = 4 x^6 y^4$

using algebraic formula $(a^m)^n= a^{mn}$

now

$2^{m} x^{mn} y^{2m} = 4x^{6} y^{4}$

now equating 'x' powers, we get

$x^{mn} = x^{6}$

$mn=6$ ....(1)

now

$y^{2m} = y^{4}$

Equating 'y' powers ,we get

2 m=4

m=2

substitute m= 2 in equation (1)

we get

2 n=6

n=3

verification:-

substitute m=2 and n=3 , we get

$[ 2 x^{n}y^{2} ]^m = 4 x^6 y^4$

$(2 x^{3} y^2)^2 = 4 x^6 y ^4\\$

$4 x^6 y^4 = 4 x^6 y^4$

both are equating so m= 2 and n=3

4 0

2 years ago

2x2 + 2x - 1 when x = -3

myrzilka [38]

2x2+2x-1=
4+2(-3)-1=
4+(-6)-1=
4+(-7)=
-3

4 0

2 years ago

Read 2 more answers

Simplify the given expression below 4/3-2i

ANTONII [103]

So-called simplifying, really means, "rationalizing the denominator", which is another way of saying, "getting rid of that pesky radical in the bottom"

$\bf \cfrac{4}{3-2i}\cdot \cfrac{3+2i}{3+2i}\impliedby \textit{multiplying by the conjugate of the bottom}
\\\\\\
\cfrac{4(3+2i)}{(3-2i)(3+2i)}\implies \cfrac{4(3+2i)}{3^2-(2i)^2}\implies \cfrac{4(3+2i)}{3^2-(4i^2)}\\\\
-------------------------------\\\\
recall\qquad i^2=-1\\\\
-------------------------------\\\\
\cfrac{4(3+2i)}{3^2-(4\cdot -1)}\implies \cfrac{4(3+2i)}{9+4}\implies \cfrac{12+8i}{13}\implies \cfrac{12}{13}+\cfrac{8}{13}i$

4 0

3 years ago

Read 2 more answers