Can someone help me please

Determine whether the graph of the quadratic function y = –x^2 – 10x + 1 opens upward or downward. Explain.

REY [17]

C. If the first coefficient is negative, then the graph opens downwards

3 0

3 years ago

A bag contains pennies, nickels, nickels, and nickels. There are 50 coins in total. Of the coins, 14% are pennies and 34% are di

Alborosie

answer:

8 pennies,16 dimes,13 nickels,13 quarters

Step-by-step explanation:

total number of coins = 50

16% are pennies

16% of 50

16/100 x 50=(16x50)/100=800/100=8

Pennies = 8

32% are dimes

32% of 50

32/100 x 50=(32x50)/100=1600/100=16

Dimes = 16

5 more nickels than pennies

Pennies = 8

nickels = 8 + 5 = 13

nickels = 13

8+16+13=37

50 - 37 = 13

13 quarters

pennies = 8, dimes = 16, nickels = 13, quarters = 13

8 0

2 years ago

Please Help Me! I need to pass.

Andru [333]

42 is the answer

Explanation:
50% of 80 is half of 80 which is 40

Then you would add 40 with 5% of 40

What you would do then is 5% times 40 then you would change the 5% into a decimal which is 0.05 and multiply it by 40 , which would get u 2

Then you would add the price with tax
40+2 = 42

6 0

2 years ago

The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is ...?

Lina20 [59]

$\frac{d}{dx}(y^2 + (xy+1)^3 = 0)
\\
\\\frac{d}{dx}y^2 + \frac{d}{dx}(xy+1)^3 = \frac{d}{dx}0
\\
\\\frac{dy}{dx}2y + \frac{d}{dx}(xy+1)\ *3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}(xy)+\frac{d}{dx}1\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}xy+x\frac{d}{dx}y+\frac{dx}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{dx}{dx}y+x\frac{dy}{dx}+\frac{dx}{dx}0\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0$
$\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(3y+3x\frac{dy}{dx}\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + 3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y') = 0
\\
\\3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y'+2yy' ) = 0
\\
\\y'(3x +6x^2y+3x^2y+2y ) = -3x^2y^3 -6xy^2-3y

$
$y' = \frac{-3x^2y^3 -6xy^2-3y}{(3x +6x^2y+3x^2y+2y )};\ x=2,y=-1
\\
\\y' = \frac{-3(2)^2(-1)^3 -6(2)(-1)^2-3(-1)}{(3(2) +6(2)^2(-1)+3(2)^2(-1)+2(-1) )}
\\
\\y' = \frac{-3(4)(-1) -6(2)(1)+3}{(6 +6(4)(-1)+3(4)(-1)-2 )}
\\
\\y' = \frac{12 -12+3}{(6 -24-12-2 )}
\\
\\y' = \frac{3}{( -32 )}$

7 0

3 years ago

What is the lmc for 6 and 14

lana [24]

Answer:

The Lcm is 42/

Step-by-step explanation:

Least common multiple (LCM) of 6 and 14 is 42.

Hope that helped.

5 0

2 years ago

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