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astraxan [27]
3 years ago
9

PLEASE HELP WITH MY MATH!

Mathematics
2 answers:
Marta_Voda [28]3 years ago
8 0
-2^6-(6)^2= -100. Hope this helps have a great day. :)
Serjik [45]3 years ago
6 0
(-2)^6 - 6^2
= 64 - 36
= 28
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A greeting card company can produce a box of cards for $7.50. If the initial investment by the company was $50,000, how many box
larisa86 [58]

Answer:

If cost of one box of card is $7.50 then the no of cards produced is 6667 cards and if cost of one box of card is $10.50 then the no of cards produced is 4762 cards

Step-by-step explanation:

The cost of one box of cards = $7.50

Initial investment = $50,000

No of box of cards produced = Initial investment / cost of one box of card

No of box of cards produced = 50,000/7.50

No of box of cards produced = 6667 cards.

if the cost of one box of cards is increased = $10.50

No of box of cards produced = Initial investment / cost of one box of card

No of box of cards produced = 50,000/10.50

No of box of cards produced = 4762 cards.

So, if cost of one box of card is $7.50 then the no of cards produced is 6667 cards and if cost of one box of card is $10.50 then the no of cards produced is 4762 cards

5 0
4 years ago
Find the sum of the constants a, h, and k such that
ra1l [238]

Answer:

  3

Step-by-step explanation:

The value of "a" is the coefficient of x^2, so we know that is 2.

__

<u>Solve for h</u>

Now, we have ...

  2x^2 -8x +7 = 2(x -h)^2 +k

Expanding the right side gives us ...

  = 2(x^2 -2hx +h^2) +k

  = 2x^2 -4hx +2h^2 +k

Comparing x-terms, we see ...

  -4hx = -8x

  h = (-8x)/(-4x) = 2

__

<u>Solve for k</u>

Now, we're left with ...

  2h^2 +k = 7 = 2(2^2) +k = 8 +k

Subtracting 8 we find k to be ...

  k = 7 -8 = -1

__

And the sum of constants a, h, and k is ...

  a +h +k = 2 +2 -1 = 3

The sum of the constants is 3.

6 0
4 years ago
(ax + b) × (cx + d) =
Stolb23 [73]
These are easy if you know  " FOIL ".  That's a procedure that takes you through multiplying two binomials.

FOIL stands for
-- <u><em>F</em></u>irst terms
<span>-- </span><u><em>O</em></u>utside terms
<span>--<em> </em></span><u><em>I</em></u>nside terms
<span>--<em> </em><u><em>L</em></u></span>ast terms

and that's how you keep everything straight while you're doing it.

(ax + b) x (cx + d)

Multiply First terms . . . 'ax' times 'cx' =  <em>acx²</em>

Multiply Outside terms . . . 'ax' times 'd'  =  <em>adx</em>

Multiply Inside terms . . .  'b' times 'cx'  =  <em>bcx</em>

Multiply Last terms . . .  'b' times 'd' =  <em>bd</em>

Now addummup:

(ax + b) x (cx + d)  =  <em>acx² + adx + bcx + bd</em>

From there, you can look for opportunities to make it look cleaner and prettier ... factoring, combining like terms, etc.

7 0
4 years ago
Read 2 more answers
Find the inverse of the function
lozanna [386]

as you already know, to get the inverse of any expression we start off by doing a quick switcheroo on the variables and then solving for "y", let's do so.

\stackrel{h(x)}{y}~~ = ~~6\sqrt[3]{2x+5}-1\implies \stackrel{\textit{quick switcheroo}}{x~~ = ~~6\sqrt[3]{2y+5}-1} \\\\\\ x+1=6\sqrt[3]{2y+5}\implies \cfrac{x+1}{6}=\sqrt[3]{2y+5}\implies \left( \cfrac{x+1}{6} \right)^3=\left( \sqrt[3]{2y+5} \right)^3

\left( \cfrac{x+1}{6} \right)^3=2y+5\implies \left( \cfrac{x+1}{6} \right)^3-5=2y\implies \cfrac{\left( \frac{x+1}{6} \right)^3-5}{2}=y \\\\\\ \cfrac{\left( \frac{x+1}{6} \right)^3}{2}-\cfrac{5}{2}=y\implies \cfrac{~~ \frac{(x+1)^3}{6^3}~~}{2}-\cfrac{5}{2}=y\implies \cfrac{(x+1)^3}{432}-\cfrac{5}{2}=\stackrel{y}{h^{-1}(x)}

4 0
2 years ago
ALMOST 20 POINTS!!!!!!!!!!
olya-2409 [2.1K]

Answer:

C

A parallelogram has 2 sets of parallel lines, but this has only one.

7 0
2 years ago
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