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GaryK [48]
3 years ago
5

B) Find the values of b. b2 = 256 b =

Mathematics
2 answers:
Svetach [21]3 years ago
5 0
B=16,-16................
Olenka [21]3 years ago
5 0

Answer:

128

Step-by-step explanation:

2*b = 256

1*b = 256 / 2 = 128

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PLS ANSWER ASAP Find the length of the side labeled x. Round intermediate values to the nearest tenth. Use the rounded values to
irina [24]
The correct answer is c. 16.6


Explain


The triangle left side
Opposite side= hypotheses x sin 0

Plugin

Opposite side = 39 x sin 43

39 x 0.68

26.60


The right side

Opposite side = adjective side x tan0

Opposite side = 26.60 x tan 32

26.60 x 0.62

X= 16.6


Hope this help you :D
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6 hm cubed to km cubed​
zvonat [6]

Answer:

0,006 I assume.

Step-by-step explanation:

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Y = x - 2<br> y = 3x + 4
faust18 [17]

Answer:

How does it need solved? By graphing? Substitution?

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3 years ago
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Consider the original complex figure and the reduction.
Roman55 [17]

Answer: The answer is C. 5

Step-by-step explanation:

Just took the test on edg.

4 0
3 years ago
Help appreciated on question in image!<br> Thanks:)
Verdich [7]

Answer:

x=-1,\:x=-7,\:x=i,\:x=-i

Step-by-step explanation:

Considering the equation

x^4+8x^3+8x^2+8x+7=0

Solving

x^4+8x^3+8x^2+8x+7

\mathrm{Factor\:}x^4+8x^3+8x^2+8x+7:\quad \left(x+1\right)\left(x+7\right)\left(x^2+1\right)

As

\mathrm{Use\:the\:rational\:root\:theorem}

a_0=7,\:\quad a_n=1

\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:7,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1

\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:7}{1}

-\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+1

=\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A]

Solving

\frac{x^4+8x^3+8x^2+8x+7}{x+1}

=x^3+7x^2+x+7

Putting \frac{x^4+8x^3+8x^2+8x+7}{x+1} =  x^3+7x^2+x+7 in equation [A]

So,

\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A]

=\left(x+1\right)x^3+7x^2+x+7

As

x^3+7x^2+x+7=\left(x+7\right)\left(x^2+1\right)

So,

Equation [A] becomes

=\left(x+1\right)\left(x+7\right)\left(x^2+1\right)

So,  the polynomial equation becomes

\left(x+1\right)\left(x+7\right)\left(x^2+1\right)=0

\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)\mathrm{Solve\:}\:x+1=0:\quad x=-1

\mathrm{Solve\:}\:x+7=0:\quad x=-7

\mathrm{Solve\:}\:x^2+1=0:\quad x=i,\:x=-i

\mathrm{The\:solutions\:are}

x=-1,\:x=-7,\:x=i,\:x=-i

Keywords: polynomial equation

Learn polynomial equation from brainly.com/question/12240569

#learnwithBrainly

5 0
3 years ago
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