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RSB [31]
3 years ago
12

When must a system of linear equations be solved algebraically, not graphically?

Mathematics
1 answer:
siniylev [52]3 years ago
4 0
<span>The correct answer is the Option "A": Integer. Then, the text is: "Whenever the solution to a system does not have integer coordinates, it can be difficult or impossible to find the precise answer through graphing."
 
 By definition, the numbers called "Integers" are all the positive numbers (which are known as</span> <span>"Natural numbers"), all the negative numbers and the zero (0) and they are identified with the letter "Z": 
 
 Z= </span>{-∞...-3,-2,-1,0,1,2,3...+∞}<span>

</span>
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Please help! will give brainlist
Alchen [17]

Answer:

C. Slope is 2

Step-by-step explanation:

Since this equation is written in the y=mx+b equation, 2 would be the slope, because any number that comes before x (AKA "m") is always the slope. For example, if you were trying to find the slope for the equation y=4x+19, 4 would be the slope.

5 0
3 years ago
1. Add or subtract. Show your work for each problem.
Aleks04 [339]
Hi there!

a. 8x^2 - 1
(x^2 - 4x + 5) + (7x^2 + 4x - 6)
= x^2 + 7x^2 - 4x + 4x - 1
= 8x^2 - 1

b. 5x^2 + 7x - 7
(7x^2 + 4x - 6) - (2x^2 - 3x + 1)
= 7x^2 + 4x - 6 - 2x^2 + 3x - 1
= 7x^2 - 2x^2 + 4x + 3x - 6 - 1
= 5x^2 + 7x -7

Hope this helps!
4 0
4 years ago
State a cubic or quartic function with the least degree possible in intercept form for the given graph. Assume that all x-interc
Iteru [2.4K]

Answer:

y = -x^{3} - 2x^{2} + 5x + 6

Step-by-step explanation:

The function will be cubic.  The x-intercepts are -4, -1, and 2

The constant factor is -1 because the graph falls on the right.  So,

y = -(x + 4)(x + 1)(x - 2)

y = -x^{3} -2x^{2} + 5x +6

6 0
3 years ago
Find the sum of the first one hundred positive integers. see fig 6.26
nignag [31]
Here's a pattern to consider:
1+100=101
2+99=101
3+98=101
4+97=101
5+96=101
.....
This question relates to the discovery of Gauss, a mathematician. He found out that if you split 100 from 1-50 and 51-100, you could add them from each end to get a sum of 101. As there are 50 sets of addition, then the total is 50×101=5050
So, the sum of the first 100 positive integers is 5050.

Quick note
We can use a formula to find out the sum of an arithmetic series:
s =  \frac{n(n + 1)}{2}
Where s is the sum of the series and n is the number of terms in the series. It works for the above problem.
8 0
3 years ago
Freeeeeeeeeeeeeeeeeeeeeeer pointk​
tigry1 [53]

welp thank you for this

7 0
3 years ago
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