What is completely factored form of d^4-81

Mathematics

1 answer:

romanna [79]3 years ago

5 0

Answer:

(d − 3) (d + 3) (d² + 9)

Step-by-step explanation:

d⁴ − 81

Factor using difference of squares.

(d² − 9) (d² + 9)

Factor using difference of squares again.

(d − 3) (d + 3) (d² + 9)

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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

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3 years ago

If a tree diagram were drawn to determine the number of possible outcomes when choosing one of 2 shirts, one of 7 ties , and one

uysha [10]

2x7x4=56

brainliest G:)))))))

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3 years ago

Read 2 more answers

How many solutions do these equations have

Arlecino [84]

Answer:

$\large\boxed{\bold{one\ solution}\ (1,\ -4)\to x=1,\ y=-4}$

Step-by-step explanation:

$\left\{\begin{array}{ccc}y=2x-6\\y=-x-3\end{array}\right\\\\\text{These are linear functions. We only need two points to draw a graph.}\\\\\text{Choice two values of x, substitute to the equation,}\\\text{and calculate the values of y.}\\\\y=2x-6\\for\ x=0\to y=2(0)-6=0-6=-6\to(0,\ -6)\\for\ x=3\to y=2(3)-6=6-6=0\to(3,\ 0)\\\\y=-x-3\\for\ x=0\to y=-0-3=0-3=-3\to(0,\ -3)\\for\ x=-3\to y=-(-3)-3=3-3=0\to(-3,\ 0)\\\\\text{Look at the picture}$