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riadik2000 [5.3K]
3 years ago
5

How do you find the x and y intercept on a table

Mathematics
1 answer:
Simora [160]3 years ago
4 0
X intercept is where x=0
y intercept is where y=0
You might be interested in
Which symbol correctly compares these fractions? −4/5 −8/9
mrs_skeptik [129]

Answer:

" > "

Step-by-step explanation:

-4/5=4x9 5×9

-4/5 = - 36/ 45

-8/9=8×5 9×5

-8/9 =-40/45

-36/ 45 > -40/45

-4/5 > -8/9

in this case - 4/5 is closer to 0

4 0
1 year ago
Read 2 more answers
Pls help with these questions
Vsevolod [243]
1. 11.32
2. 69.12
Hope it helps
4 0
3 years ago
Riley is playing a game that requires rolling a number cube numbered 1-6 and spinning a spinner with three consonants and one vo
Ghella [55]

Answer:

1/8 :)

Step-by-step explanation:

1/8 is correct because 12.5*8=100

5 0
3 years ago
How is a calculation of net worth different from a day-to-day or month-to-month tallying of expenses?
alina1380 [7]
The difference in tallying of expenses from a day to day and month to month is the consistency. in a day to day basis it is consistent the time for a day is 24 hrs it will not change, so you will really know why the expenses goes up or down/ unlike for a month to month it is inconsistent,, some months have 30, 31 or 28 days in a month.
8 0
3 years ago
Use the discriminant to find the number and type of solution for x^2+6x=-9​
likoan [24]

Answer:

D = 0; one real root

Step-by-step explanation:

Discriminant Formula:

\displaystyle \large{D =  {b}^{2}  - 4ac}

First, arrange the expression or equation in ax^2+bx+c = 0.

\displaystyle \large{ {x}^{2}  + 6x =  - 9}

Add both sides by 9.

\displaystyle \large{ {x}^{2}  + 6x + 9 =  - 9 + 9} \\  \displaystyle \large{ {x}^{2}  + 6x + 9 =  0}

Compare the coefficients so we can substitute in the formula.

\displaystyle \large{a {x}^{2}  + bx + c =  {x}^{2}  + 6x + 9 }

  • a = 1
  • b = 6
  • c = 9

Substitute a = 1, b = 6 and c = 9 in the formula.

\displaystyle \large{D =  {6}^{2}  - 4(1)(9)} \\  \displaystyle \large{D =  36 - 36} \\  \displaystyle \large{D =  0}

Since D = 0, the type of solution is one real root.

6 0
3 years ago
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