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Ne4ueva [31]
3 years ago
9

HELP

Mathematics
1 answer:
SVEN [57.7K]3 years ago
5 0

Answer:

It's Tan.

Multiple Thumbs-up ^_^

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MY RECENT QUESTION IS WORTH 47 POINTS! JUST CLICK ON MY PROFILE AND GO TO QUESTIONS TO ANSWER IT PLEASE!!!!!!!
Irina18 [472]

Answer:

ok

Step-by-step explanation:

5 0
3 years ago
If x is not equal to 0 then u/x + 5u/x - u/5x = what?
gavmur [86]
\frac{u}{x}+\frac{5u}{x}-\frac{u}{5x}=\frac{5u}{5x}+\frac{25u}{5x}-\frac{u}{5x}=\frac{5u+25u-u}{5x}=\frac{29u}{5x}
5 0
3 years ago
Read 2 more answers
Explain why f(x) = x^2+4x+3/x^2-x-2 is not continuous at x = -1.
liberstina [14]

Answer:

The value of x = -1 makes the denominator of the function equal to zero. That is why this value is not included in the domain of f(x)

Step-by-step explanation:

We have the following expression

f(x) = \frac{x^2+4x+3}{x^2-x-2}

Since the division between zero is not defined then the function f(x) can not include the values of x that make the denominator of the function zero.

Now we search that values of x make 0 the denominator factoring the polynomial x^2-x-2

We need two numbers that when adding them get as a result -1 and when multiplying those numbers, obtain -2 as a result.

These numbers are -2 and 1

Then the factors are:

(x-2) (x + 1)

We do the same with the numerator

x^2+4x+3

We need two numbers that when adding them get as a result 4 and when multiplying those numbers, obtain 3 as a result.

These numbers are 3 and 1

Then the factors are:

(x+3)(x + 1)

Therefore

f(x) = \frac{(x+3)(x+1)}{(x-2)(x+1)}

Note that \frac{(x+1)}{(x+1)}=1 only if x \neq -1

So since x = -1 is not included in the domain the function has a discontinuity in x = -1

3 0
3 years ago
Simplify the algebraic expression: (Combine like terms)<br><br> 25 + 3x + 16y - 15 + x - 10y
Maslowich
10+4x+6y because 25-15 3+1 and 16-10
4 0
3 years ago
Read 2 more answers
On a standard reticulocyte preparation with new methylene blue, there are 100 cells counted with blue-stained granulofilamentous
DIA [1.3K]

Answer:

0.322 × 10¹² /L

Step-by-step explanation:

Data provided in the question:

Number of cells counted with blue-stained granulofilamentous material

i.e number of RET = 100

The red blood cells count = 3.22 × 10¹² /L

Hematocrit = 30%

Now,

RET% = [ [ Number of RET ] ÷ 1000 RBCs ] × 100%

= [ 100 ÷ 1000 ] × 100%

= 0.1 × 100%

= 10%

also,

Absolute reticulocyte count = ( %RET × RBC count ) ÷ 100

= [ 10 × 3.22 × 10¹² /L ] ÷ 100

= 0.322 × 10¹² /L

8 0
3 years ago
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