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zepelin [54]
3 years ago
12

Find the perimeter of LMNO. (Hint: Solve for x and y first.)

Mathematics
2 answers:
Damm [24]3 years ago
7 0

6x-7=2x+9

4x=16

x=4

12=y^2+3

9=y^2

3=y -3=y

<u><em>Okay?</em></u>

Nastasia [14]3 years ago
4 0

Step-by-step explanation:

6x -7 = 2x +9

4 x = 16

X= 4

y2 +3 =12

y2 = 9

y2 =3

two side is 17 ,12

Perimeter =( 17+12 )x 2 = 58

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A spreadsheet used by accountants for Company XYZ contains 25 fields in which data is manually entered. If historical data revea
Ulleksa [173]

Answer:

See explaination for the details of the answer

Step-by-step explanation:

Defect per opportunity DPO = defects/ no. of opportunities = 2/25 = 0.08

Defect per million opportunities

DPMO = DPO * 1 million

DPMO = 0.08 * 1 million = 80,000

six sigma = 2.9, take the dpmo at higher level if not exact 80800 for 2.9

5 0
3 years ago
What is 4 radical 3 divided by 2?
Arte-miy333 [17]
3/2 = 6 
6/4= 6/40= 6 .999= ~ 7
5 0
3 years ago
Terrell de shays statement balance is $432.10 he has one outstanding check for $37.00 what is terrells adjusted balance?
KiRa [710]

Answer:

add 37.00 to 432.10. so you get 432.10 + 37.00 = 469.1

5 0
3 years ago
Simplify 7.32a+2.1∙(2.7−18a)
tresset_1 [31]
7.32a + 2.1(2.7 -18a)
7.32a + 5.67 - 37.8a (multiply 2.1 to 2.7 and 18a)
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6 0
3 years ago
Given f(x) =
sergejj [24]

Answer:

A

Step-by-step explanation:

We are given the function:

\displaystyle f(x) = \left\{        \begin{array}{ll}            2\cos(\pi x) \text{ for }  x \leq -1 \\ \\          \displaystyle   \frac{2}{\cos(\pi x)}\text{ for } x > -1        \end{array}    \right.

And we want to find:

\displaystyle \lim_{x\to -1}f(x)

So, we need to determine whether or not the limit exists. In other words, we will find the two one-sided limits.

Left-Hand Limit:

\displaystyle \lim_{x\to-1^-}f(x)

Since we are approaching from the left, we will use the first equation:

\displaystyle =\lim_{x\to -1^-}2\cos(\pi x)

By direct substitution:

=2\cos(\pi (-1))=2\cos(-\pi)=2(-1)=-2

Right-Hand Limit:

\displaystyle \lim_{x\to -1^+}f(x)

Since we are approaching from the right, we will use the second equation:

=\displaystyle \lim_{x\to -1^+}\frac{2}{\cos(\pi x)}

Direct substitution:

\displaystyle =\frac{2}{\cos(\pi (-1))}=\frac{2}{\cos(-\pi)}=\frac{2}{(-1)}=-2

So, we can see that:

\displaystyle \displaystyle \lim_{x\to-1^-}f(x)=\displaystyle \lim_{x\to -1^+}f(x) =-2

Since both the left- and right-hand limits exist and equal the same thing, we can conclude that:

\displaystyle \lim_{x \to -1}f(x)=-2

Our answer is A.

8 0
3 years ago
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