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Lilit [14]
3 years ago
10

Jack and Jill are trying to find the slope of a line segment connecting two points, (x1, y1) and (x2, y2). Jack uses the formula

m = y2 - y1 x2 - x1 . Jill mistakenly uses a different formula, but still gets the right answer. Which formula COULD she have used?
Mathematics
1 answer:
Nikolay [14]3 years ago
4 0
The correct question is
<span>Jack and Jill are trying to find the slope of a line segment connecting two points, (x1, y1) and (x2, y2). Jack uses the formula m = (y2 - y1)/( x2 - x1) . Jill mistakenly uses a different formula, but still gets the right answer. Which formula COULD she have used? A) (x2 - x1)/( y2 - y1) B) (y1 - y2)/( x1 - x2) C) (x1 - x2)/( y2 - y1) D) (y2 - y1)/( x1 - x2)

we know that

the correct formula to calculate the slope is
m=(y2-y1)/(x2-x1)
then 
</span><span>if both numerator and denominator multiply it by -1
</span>
m=-(y2-y1)/-(x2-x1)---------> m=[-y2+y1]/[-x2+x1]
m=[y1-y2]/[x1-x2]

therefore

the answer is the option 
B) (y1 - y2)/( x1 - x2)

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The solution of the following equation as 0.15 ( 10t - 9= 0.75 (4t - 3)​
Alenkinab [10]

Answer:

\huge\boxed{\sf t = 0.6}

Step-by-step explanation:

\sf 0.15(10t-9) = 0.75(4t-3)\\\\Resolving \ Parenthesis\\\\1.5t-1.35 = 3t - 2.25\\\\Combining \ like \ terms\\\\-1.35 + 2.25 = 3t-1.5t\\\\0.9 = 1.5t\\\\Dividing \ both \ sides \ by \ 1.5\\\\0.9/1.5 = t\\\\0.6 = t\\\\OR\\\\\bold{t = 0.6}

Hope this helped!

<h2>~AnonymousHelper1807</h2>
4 0
3 years ago
Hi there can anyone help me with this question please
Umnica [9.8K]
Range is 1 because they all go up by 1
8 0
3 years ago
Please help! Explain to me how you found your answer.
Katena32 [7]

Step-by-step explanation:

Sin: 2 square root 3: 4

Cos: 2:4

Tan: 2 square root 3: 2

6 0
3 years ago
Read 2 more answers
Wha bout dis one????/
PolarNik [594]
6-3=3, so m=3. Easy as pie
5 0
3 years ago
Which angle has a sine of -1/2 and a cosine of -√3/2
Sergeeva-Olga [200]
\sin x=-\dfrac{1}{2} < 0\\\\\cos x=-\dfrac{\sqrt3}{2} < 0\\\\therefore\ x\in(180^o;\ 270^o)

\text{We know:}\ \tan x=\dfrac{\sin x}{\cos x}\\\\\text{therefore:}\ \tan x=\dfrac{-\frac{1}{2}}{-\frac{\sqrt3}{2}}=\dfrac{1}{2}\cdot\dfrac{2}{\sqrt3}=\dfrac{1}{\sqrt3}\cdot\dfrac{\sqrt3}{\sqrt3}=\dfrac{\sqrt3}{3}

\tan x=\dfrac{\sqrt3}{3}\Rightarrow x=30^o+180^o\cdot k;\ k\in\mathbb{Z}

x\in(180^o;\ 270^o)\ therefore\ \boxed{x=30^o+180^o=210^o}
8 0
3 years ago
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