WhAT IS GRAPH( Y=X+3

What angle pair is pictured?

True [87]

Step-by-step explanation:

Alternate and corresponding angles.

8 0

2 years ago

Read 2 more answers

Consider the following set of vectors. v1 = 0 0 0 1 , v2 = 0 0 3 1 , v3 = 0 4 3 1 , v4 = 8 4 3 1 Let v1, v2, v3, and v4 be (colu

Alona [7]

Answer:

We have the equation

$c_1\left[\begin{array}{c}0\\0\\0\\1\end{array}\right] +c_2\left[\begin{array}{c}0\\0\\3\\1\end{array}\right] +c_3\left[\begin{array}{c}0\\4\\3\\1\end{array}\right] +c_4\left[\begin{array}{c}8\\4\\3\\1\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]$

Then, the augmented matrix of the system is

$\left[\begin{array}{cccc}0&0&0&8\\0&0&4&4\\0&3&3&3\\1&1&1&1\end{array}\right]$

We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:

$\left[\begin{array}{cccc}1&1&1&1\\0&3&3&3\\0&0&4&4\\0&0&0&8\end{array}\right]$

This matrix is in echelon form. Then, now we apply backward substitution:

1.

$8c_4=0\\c_4=0$

2.

$4c_3+4c_4=0\\4c_3+4*0=0\\c_3=0$

3.

$3c_2+3c_3+3c_4=0\\3c_2+3*0+3*0=0\\c_2=0$

4.

$c_1+c_2+c_3+c_4=0\\c_1+0+0+0=0\\c_1=0$

Then the system has unique solution that is $(c_1,c_2c_3,c_4)=(0,0,0,0)$ and this imply that the vectors $v_1,v_2,v_3,v_4$ are linear independent.

4 0

2 years ago

Solve 25^2x+1 = 144

Sergeu [11.5K]

Answer:

X = 0.272

Step-by-step explanation:

Firstly, use a common log on the 25 to undo it. Secondly, use the log25(144) on the other side to get ~ 0.54. Move the 1 over with subtraction and then divide out the 2. This will leave you with X= 0.272

4 0

3 years ago

How do I do a ratio?

Kamila [148]

A ratio is comparing one thing to another (numbers in this case). You can write ratios in many ways
here’s three different ways.
There are four girls and and 2 boys.
The ratio of girls to boys is:
4:2
4 to 2
4/2

*THE RATIO MUST ALWAYS BE IN ORDER!!! If boys come first in the problem WRITE THE BOYS NUMBER FIRST! If girls come first, WRITE THE GIRLS FIRST IN THE RATIO!!! (That’s a very common mistake) hope this helps!

6 0

2 years ago

Can I get help solving this graph please?

Daniel [21]

see the attached figure with the letters

1) find m(x) in the interval A,B
A (0,100) B(50,40) -------------- > p=(y2-y1(/(x2-x1)=(40-100)/(50-0)=-6/5
m=px+b---------- > 100=(-6/5)*0 +b------------- > b=100
mAB=(-6/5)x+100

2) find m(x) in the interval B,C
B(50,40) C(100,100) -------------- > p=(y2-y1(/(x2-x1)=(100-40)/(100-50)=6/5
m=px+b---------- > 40=(6/5)*50 +b------------- > b=-20
mBC=(6/5)x-20

3) find n(x) in the interval A,B
A (0,0) B(50,60) -------------- > p=(y2-y1(/(x2-x1)=(60)/(50)=6/5
n=px+b---------- > 0=(6/5)*0 +b------------- > b=0
nAB=(6/5)x

4) find n(x) in the interval B,C
B(50,60) C(100,90) -------------- > p=(y2-y1(/(x2-x1)=(90-60)/(100-50)=3/5
n=px+b---------- > 60=(3/5)*50 +b------------- > b=30
nBC=(3/5)x+30

5) find h(x) = n(m(x)) in the interval A,B
mAB=(-6/5)x+100
nAB=(6/5)x
then
n(m(x))=(6/5)*[(-6/5)x+100]=(-36/25)x+120
h(x)=(-36/25)x+120
find h'(x)
h'(x)=-36/25=-1.44

6) find h(x) = n(m(x)) in the interval B,C
mBC=(6/5)x-20
nBC=(3/5)x+30
then
n(m(x))=(3/5)*[(6/5)x-20]+30 =(18/25)x-12+30=(18/25)x+18
h(x)=(18/25)x+18
find h'(x)
h'(x)=18/25=0.72

for the interval (A,B) h'(x)=-1.44
for the interval (B,C) h'(x)= 0.72

 h'(x) = 1.44 ------------ > not exist

8 0

2 years ago