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3 years ago

HELP

Mathematics

1 answer:

SVEN [57.7K]3 years ago

5 0

Answer:

It's Tan.

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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:

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