Answer:
Let us take 'a' in the place of 'y' so the equation becomes
(y+x) (ax+b)
Step-by-step explanation:
Step 1:
(a + x) (ax + b)
Step 2: Proof
Checking polynomial identity.
(ax+b )(x+a) = FOIL
(ax+b)(x+a)
ax^2+a^2x is the First Term in the FOIL
ax^2 + a^2x + bx + ab
(ax+b)(x+a)+bx+ab is the Second Term in the FOIL
Add both expressions together from First and Second Term
= ax^2 + a^2x + bx + ab
Step 3: Proof
(ax+b)(x+a) = ax^2 + a^2x + bx + ab
Identity is Found .
Trying with numbers now
(ax+b)(x+a) = ax^2 + a^2x + bx + ab
((2*5)+8)(5+2) =(2*5^2)+(2^2*5)+(8*5)+(2*8)
((10)+8)(7) =(2*25)+(4*5)+(40)+(16)
(18)(7) =(50)+(20)+(56)
126 =126
In the problem 56x31, what are the partial products? To acquire the possible partial products we can just multiply the two numbers to produce the possible numbers at hand. <span><span>
1. </span>56 x 31 = 1736</span><span><span>
2. </span>31 x 56 = 1736</span><span> </span>
<span>Same outcome which is explained by the comutative property of multiplication.</span>
Answer:
○ 
Explanation:
Accourding to one of the circle equations, 
the centre of the circle is represented by 
Moreover, all negative symbols give you the OPPOCITE TERMS OF WHAT THEY <em>REALLY</em> ARE, so you must pay cloce attention to which term gets which symbol. Another thing you need to know is that the radius will ALWAYS be squared, so no matter how your equation comes about, make sure that the radius is squared. Now, in case you did not know how to define the radius, you can choose between either method:
Pythagorean Theorem
Sinse we are dealing with <em>length</em>, we only desire the NON-NEGATIVE root.
Distanse Equation
![\displaystyle \sqrt{[-x_1 + x_2]^2 + [-y_1 + y_2]^2} = d \\ \\ \sqrt{[-7 + 4]^2 + [-2 - 2]^2} = r \hookrightarrow \sqrt{[-3]^2 + [-4]^2} = r \hookrightarrow \sqrt{9 + 16} = r; \sqrt{25} = r \\ \\ \boxed{5 = r}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Csqrt%7B%5B-x_1%20%2B%20x_2%5D%5E2%20%2B%20%5B-y_1%20%2B%20y_2%5D%5E2%7D%20%3D%20d%20%5C%5C%20%5C%5C%20%5Csqrt%7B%5B-7%20%2B%204%5D%5E2%20%2B%20%5B-2%20-%202%5D%5E2%7D%20%3D%20r%20%5Chookrightarrow%20%5Csqrt%7B%5B-3%5D%5E2%20%2B%20%5B-4%5D%5E2%7D%20%3D%20r%20%5Chookrightarrow%20%5Csqrt%7B9%20%2B%2016%7D%20%3D%20r%3B%20%5Csqrt%7B25%7D%20%3D%20r%20%5C%5C%20%5C%5C%20%5Cboxed%7B5%20%3D%20r%7D)
Sinse we are dealing with <em>distanse</em>, we only desire the NON-NEGATIVE root.
I am joyous to assist you at any time.
Answer:
A scientist adds drops of liquid to a test tube. The test tube has marks every ml.
Each drop contains 0.14 mL. Between which two marks on the test tube will the
liquid be after the sixth drop is added?
a. 1/5 and 2/5, between 0.2 and 0.4 mL's
b.2/5 and 3/5, between 0.4 and 0.6 ml's
c.3/5 and 4/5, between 0.6 and 0.8 mL's
d. 4/5 and 5/5, between 0.8 and 1 ml
Answer:
Graph #1 - y=-20x+40
Graph #2 - y=5x+10
Step-by-step explanation: The first thing that you do is pull two points from each of the graphs and find the slope. To find the slope you use
<h2>
M = y2-y1/x2-x1</h2>
<u>FROM GRAPH #1:</u>
<h3>(3,40) (2,80)</h3>
(x1,y1) (x2,y2)
80-40/2-3
20/-1
M = -20
<u>FROM GRAPH #2:</u>
<h3>
(1,10) (3,20)</h3>
(x1,y1) (x2,y2)
20-10/3-1
10/2
M = 5
2. Now once you find the slope you will use the equation of a line formula to form an equation.
<h2>
y=mx+b</h2>
The (mx) is your slope, the (b) is the y-intercept. (any value on the y-axis)
3. Since we now know the equation of a line and the slope we will substitute the variables.
<u>Graph #1:</u>
y=mx+b
y=-20x+40
<u>Graph #2:</u>
y=mx+b
y=5x+10
<h3>
HOPE THIS HELPS :)</h3>
