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3 years ago

I will mark brainliest

Mathematics

1 answer:

RoseWind [281]3 years ago

4 0

Answer: 56(32)+2 = 43

Step-by-step explanation:

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You do not need to be concerned if there are limits on what a company will pay for a service of medical or dental procedures.

riadik2000 [5.3K]

False because you should always make sure a company has medical and detal procedures

8 0

3 years ago

Evaluate (2 - 5i)(p + q)i when p = 2 and q = 5i<br>a. 29i<br>b. 291-20<br>C.-21i<br>d.29

attashe74 [19]

Answer:

A. 29i

Step-by-step explanation:

Step 1: Plug in given variables

(2 - 5i)(2 + 5i)i

Step 2: Difference of squares (expand)

(4 - 25i²)i

Step 3: Imaginary numbers rules

(4 - 25(-1))i

Step 4: Combine like terms

(4 + 25)i

(29)i

Your final answer will be 29i

8 0

3 years ago

Read 2 more answers

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

3 years ago

Read 2 more answers

Use the distributive property to write an expression that is equivalent to each given property.

Nadya [2.5K]

Answer:

-6x+12

-9+5a

7x+63

8y-4/5x-12

Step-by-step explanation:

To solve these all you have to do is take the number outside of the parenthesis and multiply them by each term.

-3(2x)-3(-4) All I did here was expand the problem to show what i mean by terms.

-6x+12

Do this for each problem.

0.1(-90)+0.1(50a)

-9+5a

-7(-x)-7(-9) For this one you are taking negatives times negatives so each answer will be positive.

7x+63

4/5(10y)+4/5(-x)+4/5(-15)

8y-4/5x-12

3 0

3 years ago

Please help me! And please explain how to solve this. Especially questions B, C, and E

kati45 [8]

A. radical 76 is8.6 so false
B. 3 radical 10 is 9.4 true
C. 4 radical 2 is 5.6 so false
D. radical 29 is 5.3 so true
E. 2 radical 50 is 14.14 so false

5 0

3 years ago