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3 years ago

The slope of a line is?

Mathematics

1 answer:

valkas [14]3 years ago

6 0

Answer:

The correct answer is d) the ratio of the change in y to the change in x along the line.

We can tell this by looking at the slope formula.

m (slope) = (y2 - y1)/(x2 - x1)

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Abc=3x+5 cbd=2x+5 find the following m

Monica [59]

Sry idk wish I could know but good luck

4 0

3 years ago

Solve for z.<br> 7z +4 – 9z = 8<br> z = [?]

ZanzabumX [31]

Answer:

Z = -2

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

If a point at (3,-4) is reflected over the X-axis to form a new point What will be the coordinate of the new point? (-3,-4), (3,

ira [324]

Answer:

(3,4) because only the y would change when it is being reflected over the x axis :)

Step-by-step explanation:

5 0

3 years ago

Solve the system:<br> 5x-7y=-41<br> -3x-5y=-3

-Dominant- [34]

Answer:

x = -4 , y = 3

Step-by-step explanation:

5x - 7y = -41 ... (i)

-3x - 5y = -3 ... (ii)

Multiplying (i) by -3 and (ii) by 5 ;

-15x + 21y = 123 ... (i)

-15x - 25y = -15 ... (ii)

Subtracting (i) by (ii) ;

0 + 46y = 138

46y = 138

y = 138 ÷ 46 = 3

Returning to equation (ii) ;

-3x - 5(3) = -3

-3x = -3 + 15

-3x = 12

x = -4

8 0

3 years ago

Read 2 more answers

Will someone help with this ! I was allowed a retry for it and I don’t know how to do it

pashok25 [27]

Answer:

$\frac{dN}{dt}=k(725-N)\\separating the variables and integrating\\\int \frac{dN}{725-N}=\int kdt+c\\-log (725-N)=kt+c\\log (725-N)=-kt-c\\(725-N)=e^{-kt-c} =e^{-kt} *e^{-c} =Ce^{-kt} \\when t=0,N=400\\725-400=Ce^{0} =C\\C=325\\when t=3,N=650\\725-650=325e^{-3k} \\\frac{75}{325}=(e^{-k})^3\\\\e^{-k} =(\frac{3}{13}) ^{\frac{1}{3} } \\when t=5\\725-N=325(\frac{3}{13}) ^{\frac{5}{3} } =325*\frac{3}{13}*(\frac{3}{13} )^{\frac{2}{3}}\\N=725-75*(\frac{3}{13} )^{\frac{2}{3} }=725-28=697$

Step-by-step explanation:

6 0

3 years ago