What statement describes the relationship between ∆ XYZ and ∆ X'Y'Z'?

Mathematics

1 answer:

frutty [35]2 years ago

3 0

I think you forgot to add the attachment.

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HELPP!!!!!

Alja [10]

Answer:

choice b

Step-by-step explanation:

3.49

____ = 0.1396 ≈ 0.14 or 14 cents per ounce

4.79

____ = 0.10644444 ≈ 0.11 or 11 cents per ounce

3 0

2 years ago

In a batch of 890 calculators, 12 were found to be defective. What is the probability that a calculator chosen at random will be

scoray [572]

Answer:

When entered into a calculator, 12/890 (the chance of a given calculator being defective) = 0.01348...

As a percent this is 100*0.01348% = 1.348%, and to the nearest tenth, 1.3%

3 0

2 years ago

Read 2 more answers

After traveling 345 miles, the

vfiekz [6]

<h2>GOOD LUCK CHUCK :)))))))))</h2>

5 0

3 years ago

Using these complex zeros (1,1,-1/2,2+i,2-i) factor f(x)=-2x^5 +11x^4 -22x^3 +14x^2 +4x -5

Marianna [84]

$\bf \begin{cases}
x=1\implies &x-1=0\\
x=1\implies &x-1=0\\
x=-\frac{1}{2}\implies 2x=-1\implies &2x+1=0\\
x=2+i\implies &x-2-i=0\\
x=2-i\implies &x-2+i=0
\end{cases}
\\\\\\
(x-1)(x-1)(2x+1)(x-2-i)(x-2+i)=\stackrel{original~polynomial}{0}
\\\\\\
(x-1)^2(2x+1)~\stackrel{\textit{difference of squares}}{[(x-2)-(i)][(x-2)+(i)]}$

$\bf (x^2-2x+1)(2x+1)~[(x-2)^2-(i)^2]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)-(-1)]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)+1]
\\\\\\
(x^2-2x+1)(2x+1)~[x^2-4x+5]
\\\\\\
(x^2-2x+1)(2x+1)(x^2-4x+5)$

of course, you can always use (x-1)(x-1)(2x+1)(x²-4x+5) as well.

7 0

2 years ago

What is the value of x?<br><br> Enter your answer, as a decimal, in the box.

Katarina [22]

Answer:

$x=50.6cm$