1. 4x754=5278
2. you could use exponets
Answer: false
Step-by-step explanation: dunno
Answer:
Please check the attached figure!
Step-by-step explanation:
Part a)
Point A is located at the x-coordinate x=-4 and y-coordinate y=1.
Hence, the coordinates of point A = (-4, 1)
Part b)
Point B(3, -2) has been plotted and is shown in the attached figure.
It is clear from the attached diagram that point B is located at the x-coordinate x=3 and y-coordinate y=-2. Hence, the coordinates of point B = (3, -2)
Part C)
Point C has the same x-coordinate as point A i.e. x=-4 and the same y-coordinate as point B i.e. y=-2.
Hence, the coordinates of point C = (-4, -2). Point C is also plotted as shown in the diagram.
Answer:
A solution is said to be extraneous, if it is a zero of the equation, but it does not satisfy the equation,when substituted in the original equation,L.H.S≠R.H.S.
The given equation consisting of variable , m is
![\frac{2 m}{2 m+3} -\frac{2 m}{2 m-3}=1\\\\ 2 m[\frac{1}{2 m+3} -\frac{1}{2 m-3}]=1\\\\ 2 m\times \frac{[2 m-3 -2 m- 3]}{4m^2-9}=1\\\\ -6 \times 2 m=4 m^2 -9\\\\ 4 m^2 +1 2 m -9=0\\\\m=\frac{-12 \pm\sqrt{12^2-4 \times 4 \times (-9)}}{2\times 4}\\\\m=\frac{-12 \pm \sqrt {144+144}}{8}\\\\m=\frac{-12 \pm \sqrt {288}}{8}\\\\m=\frac{-12 \pm 12 \sqrt{2}}{8}\\\\m=\frac{3}{2}\times(-1 \pm \sqrt{2})](https://tex.z-dn.net/?f=%5Cfrac%7B2%20m%7D%7B2%20m%2B3%7D%20-%5Cfrac%7B2%20m%7D%7B2%20m-3%7D%3D1%5C%5C%5C%5C%202%20m%5B%5Cfrac%7B1%7D%7B2%20m%2B3%7D%20-%5Cfrac%7B1%7D%7B2%20m-3%7D%5D%3D1%5C%5C%5C%5C%202%20m%5Ctimes%20%5Cfrac%7B%5B2%20m-3%20-2%20m-%203%5D%7D%7B4m%5E2-9%7D%3D1%5C%5C%5C%5C%20-6%20%5Ctimes%202%20m%3D4%20m%5E2%20-9%5C%5C%5C%5C%204%20m%5E2%20%2B1%202%20m%20-9%3D0%5C%5C%5C%5Cm%3D%5Cfrac%7B-12%20%5Cpm%5Csqrt%7B12%5E2-4%20%5Ctimes%204%20%5Ctimes%20%28-9%29%7D%7D%7B2%5Ctimes%204%7D%5C%5C%5C%5Cm%3D%5Cfrac%7B-12%20%5Cpm%20%5Csqrt%20%7B144%2B144%7D%7D%7B8%7D%5C%5C%5C%5Cm%3D%5Cfrac%7B-12%20%5Cpm%20%5Csqrt%20%7B288%7D%7D%7B8%7D%5C%5C%5C%5Cm%3D%5Cfrac%7B-12%20%5Cpm%2012%20%5Csqrt%7B2%7D%7D%7B8%7D%5C%5C%5C%5Cm%3D%5Cfrac%7B3%7D%7B2%7D%5Ctimes%28-1%20%5Cpm%20%5Csqrt%7B2%7D%29)
None of the two solution
, is extraneous.
Here, L.H.S= R.H.S
Option A: 0→ extraneous
Answer:
Step-by-step explanation:
11) First write in decreasing exponential terms and fill in blanks with zeros.
Our goal is to eliminate all term in the dividend by subtraction
________________
2v - 2 | 2v³ - 16v² + 0v + 13
we see that 2v needs to be multiplied by 1v² to eliminate the first term
<u> v² </u>
2v - 2 | 2v³ - 16v² + 0v + 13
<u>- (2v³ - 2v²)</u>
0 - 14v²
multiply your estimate by your divisor and subtract from the dividend.
bring down the next term and repeat.
<u> v² -7v </u>
2v - 2 | 2v³ - 16v² + 0v + 13
<u>- (2v³ - 2v²)</u>
0 - 14v² + 0v
<u>-(-14v² + 14v)</u>
- 14v
repeat again
<u />
<u> v² - 7v - 7 </u>
2v - 2 | 2v³ - 16v² + 0v + 13
<u>- (2v³ - 2v²)</u>
0 - 14v² + 0v
<u>-(-14v² + 14v)</u>
- 14v + 13
<u>-(-14v + 14)</u>
-1
and remainder gets put over the divisor and appended
v² - 7v - 7 - 1/(2v - 2)
13) Same process
<u> </u>
5a + 1 | 40a³ - 12a² - 39a - 5
<u> 8a² </u>
5a + 1 | 40a³ - 12a² - 39a - 5
<u>-(40a³ + 8a²)</u>
-20a² - 39a
<u> 8a² - 4a </u>
5a + 1 | 40a³ - 12a² - 39a - 5
<u>-(40a³ + 8a²)</u>
-20a² - 39a
-(-<u>20a² - 4a)</u>
-35a - 5
<u> 8a² - 4a - 7 </u>
5a + 1 | 40a³ - 12a² - 39a - 5
<u>-(40a³ + 8a²)</u>
-20a² - 39a
-(<u>20a² - 4a)</u>
-35a - 5
-<u>(-35a - 7)</u>
2
8a² - 4a - 7 + 2/(5a + 1)
