Answer:
Step-by-step explanation:
The ratio of corresponding sides DN and KI is 12 : 4 = 3 : 1. The same ratio applies to altitudes DQ and KO. Since the difference between these altitudes is 6 and the difference between their ratio units is 3-1 = 2, each ratio unit must stand for 6/2 = 3 units of linear measure. That is, ...
DQ = (3 units)·3 = 9 units
KO = (3 units)·1 = 3 units
Then the base lengths QN and OI can be found from the Pythagorean theorem:
KI² = KO² +OI²
4² = 3² +OI²
OI = √(16 -9)
OI = √7
QN = 3·OI = 3√7
(1,1)
(2,3)
(0,-1)
(3,5)
(4,7)
it’s actually very simple, all you have to do is plug in whatever value you want for x and then solve the problem. once you have done that you put it in this format: (x,y)
for example, the first ordered pair that i told you: (1,1)
i got this by first deciding to use 1 as my x-value, then plugging it into the equation, so it was y=2•1-1. from there you just solve for y, which is very simple. make sure you remember order of operations though, or you could get it wrong! from that step i just put it in ordered pair format, or (x,y). this made my first ordered pair, which was (1,1). you can use the same steps for all ordered pairs you want to find. i hope this helps!
A basic algebraic linear equation is just a fancy way of saying "equation of a line". They are most likely referring to the formula y=mx+b, where m is the slope and b in the y intercept. :)
They are both right because the first expression is the perimeter but expanded in terms of length.The second expression is shortened to be a factor expression but they both represent the length.
The benefit of the first one is you can see how it’s expanded and plug the numbers in and the second one is more easy to plug in the numbers with ease.
Answer:
<em>We disagree with Zach and Delia and agree with Alicia</em>
Step-by-step explanation:
The domain of a function is the set of values of the independent variable that the function can take according to given rules or restrictions.
The range is the set of values the dependent variable can take for every possible value of the domain.
The graph shows a continuous line representing the values of the function. We must take a careful look to the values of x (horizontal axis) where the function exists. It can be done by drawing an imaginary vertical line passing through the value of x. If that line touches the graph of the function, it belongs to the domain. It's clear that every value of x between -5 and 3 (both inclusive because there are solid dots in the extremes) belong to the domain:
Domain: 
The range is obtained in a similar way as the domain, but the imaginary lines must be horizontal. That gives us the values of y range from -7 to 5 both inclusive:
Range:
Thus we disagree with Zach and Delia and agree with Alicia
