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3 years ago

How to find the slope with only the x and y intercepts?

1 answer:

8 0

If we have the equation of a line: y=ax+b
How to find a if with only the x-intercepts (called m) and y-intercepts (called n)
+ First we have y=0 and x=m or 0=a*m+b or b= -a*m
+ Then we have x=0, y=n or n=a*0+b or b=n
And if m≠0 we have a= -n/m
and if m=0 this line has no slope

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HELP HELP HELP HELP WOPFOPOPMPOMGOPRDAGS

Basile [38]

Answer:

Step-by-step explanation:

You can't ever let this go negative. At least not at the grade you are in.

It can be 0.

So the domain must start at x = 7

sqrt(5*7 - 35) = sqrt(0) = 0

x can have any value (including 7) between 7 and infinity. If you choose a number less than 7 (like 6) the square root will go negative and that's not to be done.

So the interval is

7 ≤ x < ∞

4 0

2 years ago

Araceli had 61 pennies and then lost 12. Now she wants to buy a treat that costs 4 sevenths of the pennies she has left. How man

igor_vitrenko [27]

The treat would cost 7 cents of your 49

8 0

3 years ago

Efficiency is the ratio of output work to input work, expressed as a percentage. Light bulbs put out less light energy than the

grandymaker [24]

Answer:

10%

Step-by-step explanation:

Using the given formula with the given data, we have ...

efficiency = output work / input work

= (10 J)/(100 J) = 0.10 = 10%

7 0

3 years ago

A water tank is 3/4 full.

guapka [62]

x/.75=x-56/0.4

0.4x=.75x-42

-0.35x=-42

x=120

I set up a proportion in which x=the tank when it is 3/4 full. When the tank is 3/4 full, solving the proportion tells us the x=120 litres. 120/3=40, and 120+40 is 160, showing that a tank completely full would indeed hold 160 litres.

Hope this helps!

3 0

2 years ago

Which statement describes if there is an extraneous solution to the equation √x-3 = x-5? A. there are no solutions to the equati

Brrunno [24]

Remember that an extraneous solution of an equation, is the solution that emerges from solving the equation but is not a valid solution.

Lets solve our equation to find out what is the extraneous solution:
 $\sqrt{x-3} =x-5$
$(\sqrt{x-3})^2 =(x-5)^2$
$x-3=x^2-10x+25$
$x^2-11x+28=0$
$(x-4)(x-7)=0$
$x-4=0$ and $x-7=0$
$x=4$ and $x=7$

So, the solutions of our equation are $x=4$ and $x=7$ . Lets replace each solution in our original equation to check if they are valid solutions:
- For $x=7$
$\sqrt{x-3} =x-5$
$\sqrt{7-3} =7-5$
$\sqrt{4} =2$
$2=2$
We can conclude that 7 is a valid solution of the equation.

- For $x=4$
$\sqrt{x-3} =x-5$
$\sqrt{4-3} =4-5$
$\sqrt{1} =1$
$1 \neq 1$
We can conclude that 4 is not a valid solution of the equation; therefore, 4 is a extraneous solution.

We can conclude that the correct answer is: D. the extraneous solution is x = 4

7 0

2 years ago