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defon
3 years ago
13

INPUT SU,0 IF OFF OUTPUT M,0 OUTPUT T,0 OUTPUT W,0 OUTPUT TH,0 OUTPUT F,0 ELSE ON OUTPUT M,0 OUTPUT T,0 OUTPUT W,0 OUTPUT TH,0 O

UTPUT F,0 ENDIF INPUT M,0 INPUT T,0 INPUT W,0 INPUT TH,0 INPUT F,0 OR OR OR OR INPUT SA,0 INPUT SU,0 AND NOT OR ON OUTPUT SU,0 OFF OUTPUT SU,0 END
Mathematics
1 answer:
Karo-lina-s [1.5K]3 years ago
3 0
Why’s it in all caps and what’s the question this doesn’t look like a math equation
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Need help on problem number 3 and 4 pzzzz
saul85 [17]
Numbers 3 & 4 are D, let me know if you want me to explain
7 0
4 years ago
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Ryan earns $5 for each small dog he walks and $8 for each large dog he walks. Today he walked 8 dogs and made a total of $55. Ho
Sunny_sXe [5.5K]
Let L and S be the number of large and small dogs respectively.

L+S=8 so we can say L=8-S

5S+8L=55, using L found above in this equation yields:

5S+8(8-S)=55

5S+64-8S=55

-3S+64=55

-3S=-9

S=3, since L=8-S

L=5

So Ryan walked 3 small dogs and 5 large dogs.
5 0
3 years ago
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Which ordered pair is the solution to the system of equations? {x+3y=-12, y= 1/3x-6?
Julli [10]

For this, I'll be using the substitution method. Since y = 1/3x - 6, replace the y variable in the first equation with 1/3x - 6 and solve for x:

x+3(\frac{1}{3}x-6)=-12\\ x+x-18=-12\\ 2x-18=-12\\ 2x=6\\ x=3

Now that we have the value of x, plug it into either equation to solve for y:

3+3y=-12\\ 3y=-15\\ y=-5\\ \\ y=\frac{1}{3}*3-6\\ y=1-6\\ y=-5

<h3><u>In short, the solution to this equation is (3,-5).</u></h3>
5 0
3 years ago
I’m honestly confused I will mark Brainliest
Mariana [72]

Answer:

28 units !

brainliest please.

Step-by-step explanation:

6 0
3 years ago
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Find the derivative of f(x) = x/6 at x = -2.
AfilCa [17]
f'(x_0)=\lim\limits_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}\\\\f(x)=\dfrac{x}{6};\ x_0=-2\\\\subtitute\\\\f'(-2)=\lim\limits_{x\to-2}\dfrac{\frac{x}{6}-\frac{-2}{6}}{x-(-2)}=\lim\limits_{x\to-2}\dfrac{\frac{x}{6}+\frac{2}{6}}{x+2}=\lim\limits_{x\to-2}\dfrac{\frac{x+2}{6}}{x+2}\\\\=\lim\limits_{x\to-2}\left(\dfrac{x+2}{6}\cdot\dfrac{1}{x+2}\right)=\lim\limits_{x\to-2}\dfrac{1}{6}=\dfrac{1}{6}
6 0
3 years ago
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