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11111nata11111 [884]
3 years ago
15

Which is the upper left quadrant on the coordinate plane?

Mathematics
2 answers:
Jlenok [28]3 years ago
8 0

Answer:

its Quadrant IV

Step-its Quadrant by-step explanation:

Lisa [10]3 years ago
4 0

Answer:

the upper left Quadrant is Quadrant II

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What is the equation of the line through (-6, -5) and (-4, -4)?
Norma-Jean [14]

Answer:

  y = 1/2x -2

Step-by-step explanation:

The 2-point form of the equation for a line is useful for answering questions like this.

  y = (y2 -y1)/(x2 -x1)(x -x1) +y1 . . equation of a line through (x1, y1) and (x2, y2)

Filling in the given points, the equation becomes ...

  y = (-4 -(-5))/(-4 -(-6))/(x -(-6)) -5

  y = 1/2(x +6) -5

  y = 1/2x -2

3 0
3 years ago
Which equation can be used to represent "three minus the difference of a number and one equals one-half of the difference of thr
NNADVOKAT [17]
<span>3 – (n – 1) = 1/2(3n – 4)

</span>
4 0
3 years ago
Read 2 more answers
(secx)dy/dx=e^(y+sinx), please help me solve the differential equation. Thanks :)
nalin [4]
First, you must know these formula  d(e^f(x) = f'(x)e^x dx, e^a+b=e^a.e^b, and d(sinx) = cosxdx, secx = 1/ cosx

(secx)dy/dx=e^(y+sinx), implies  <span>dy/dx=cosx .e^(y+sinx), and then 
</span>dy=cosx .e^(y+sinx).dx, integdy=integ(cosx .e^(y+sinx).dx, equivalent of 
integdy=integ(cosx .e^y.e^sinx)dx, integdy=e^y.integ.(cosx e^sinx)dx, but we know that   d(e^sinx) =cosx e^sinx dx,
so integ.d(e^sinx) =integ.cosx e^sinx dx,
and e^sinx + C=integ.cosx e^sinxdx
 finally, integdy=e^y.integ.(cosx e^sinx)dx=e^2. (e^sinx) +C
the answer is 
y = e^2. (e^sinx) +C, you can check this answer to calculate dy/dx
7 0
3 years ago
Read 2 more answers
PLEASE HELP! WILL GIVE BRAINLIEST!!!!!
Sophie [7]

The 8 percent commission must be at least $1000.

8% = $1000

1% = $125

100% = $12500

Hence, Jim needs to sell $12500 dollars worth of widgets.

6 0
3 years ago
The hydrogen ion concentration (h+) in a certain cleaning compound is (h+)=3.2x10^-11 use the formula ph=-log(h+) to find the pH
Furkat [3]

Answer:

The pH of the clean compound is approximately 10.495.

Step-by-step explanation:

If the cleaning ion concentration of the cleaning compound is [H^{+}] = 3.2\times 10^{-11}, then the pH, no unit, of the compound is:

pH = -\log_{10} [H^{+}]  (1)

pH = -\log_{10} (3.2\times 10^{-11})

pH = -\log_{10} 3.2 - \log_{10} (10^{-11})

pH = -\log_{10} \frac{32}{10} +11

pH = -\log_{10} 32 + \log_{10} 10 + 11

pH = -\log_{10} 32 +12

pH \approx 10.495

The pH of the clean compound is approximately 10.495.

6 0
3 years ago
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