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erastova [34]
2 years ago
9

Simplify the expression below as much as possible. (8 + 9i) + (5 - 91) - (8 - 7i)

Mathematics
1 answer:
Sauron [17]2 years ago
7 0

Answer:

B. 5 + 7i

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Distributive Property

<u>Algebra I</u>

  • Combining Like Terms

<u>Algebra II</u>

  • Complex Standard Form: a + bi
  • Imaginary Numbers

Step-by-step explanation:

<u>Step 1: Define expression</u>

(8 + 9i) + (5 - 9i) - (8 - 7i)

<u>Step 2: Simplify</u>

  1. Distribute negative:                     8 + 9i + 5 - 9i - 8 + 7i
  2. Combine like terms:                    5 + 7i
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2) Evaluate the given numerical expression.<br><br> A) 9<br> B) 12<br> C) 14<br> D) 21
eduard

Answer:

  • \boxed{\sf{14}}

Step-by-step explanation:

\boxed{\underline{\text{ORDER OF OPERATIONS:}}}

Use the order of operations.

Note: PEMDAS or BODMAS stands for:

<u>PEMDAS</u>

  • <u>Parentheses</u>
  • <u>Exponents</u>
  • <u>Multiply</u>
  • <u>Divide</u>
  • <u>Add</u>
  • <u>Subtract</u>

<u>BODMAS</u>

  • <u>Brackets</u>
  • <u>Order</u>
  • <u>Divide</u>
  • <u>Add</u>
  • <u>Subtract</u>

<u>First, do parentheses.</u>

3+6*(5+4)÷3-7

(5+4)=9

\Longrightarrow: \sf{3+6* 9\div 3-7}

<u>Do multiply and divide.</u>

6*9=54

54/3=18

<u>Then, rewrite the problem down.</u>

\Longrightarrow: \sf{3+18-7}

<u>Add.</u>

\sf{3+18=21}

\sf{21-7=\boxed{\sf{14}}

\Longrightarrow: \boxed{\sf{14}}

  • <u>Therefore, the correct answer is "C. 14".</u>

I hope this helps, let me know if you have any questions.

8 0
2 years ago
8,000 of 2ounces how many make 16 ounces
yan [13]
1,000 I guess I don't really understand the question
HAPPY THANKSGIVING!
7 0
3 years ago
Read 2 more answers
jackson bought 555 ounces of raisins for \$4$4dollar sign, 4. How much do raisins cost per ounce? \$$dollar sign How many ounces
irga5000 [103]

Answer:

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Prove that the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus.
sergiy2304 [10]

Answer:

See explanation

Step-by-step explanation:

a) To prove that DEFG is a rhombus, it is sufficient to prove that:

  1. All the sides of the rhombus are congruent:  |DG|\cong |GF| \cong |EF| \cong |DE|
  2. The diagonals are perpendicular

Using the distance formula; d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

|DG|=\sqrt{(0-(-a-b))^2+(0-c)^2}

\implies |DG|=\sqrt{a^2+b^2+c^2+2ab}

|GF|=\sqrt{((a+b)-0)^2+(c-0)^2}

\implies |GF|=\sqrt{a^2+b^2+c^2+2ab}

|EF|=\sqrt{((a+b)-0)^2+(c-2c)^2}

\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}

|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}

\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}

Using the slope formula; m=\frac{y_2-y_1}{x_2-x_1}

The slope of EG is m_{EG}=\frac{2c-0}{0-0}

\implies m_{EG}=\frac{2c}{0}

The slope of EG is undefined hence it is a vertical line.

The slope of  DF is m_{DF}=\frac{c-c}{a+b-(-a-b)}

\implies m_{DF}=\frac{0}{2a+2b)}=0

The slope of DF is zero, hence it is a horizontal line.

A horizontal line meets a vertical line at 90 degrees.

Conclusion:

Since |DG|\cong |GF| \cong |EF| \cong |DE| and DF \perp FG , DEFG is a rhombus

b) Using the slope formula:

The slope of DE is m_{DE}=\frac{2c-c}{0-(-a-b)}

m_{DE}=\frac{c}{a+b)}

The slope of FG is m_{FG}=\frac{c-0}{a+b-0}

\implies m_{FG}=\frac{c}{a+b}

5 0
3 years ago
cos (x/2) = -sqrt2/2 in (0,360) looking for two values ?? Can someone please help me I can’t seem to figure out how to solve thi
ValentinkaMS [17]

Answer:

135° and 225°

Step-by-step explanation:

basically you want to find the value of x between 0 and 360 in this equation

cos x/2 = -(√2)/2

assume x/2 as n, so

cos n = -(√2)/2

n = 45°

then remember the quadrant system

0-90 1st quadrant, all is POSITIVE

90-180 2nd quadrant, only SIN has positive value

180 - 270 3rd quadrant, only TAN has positive value

270 -360 4th quadrant, only COS positive here.

so if you try to find negative value look into 2nd and 3rd quadrant that related 45° to x-axis (0°or 180°)

so the value of x is

180 - 45 = 135° (2nd quadrant) and

180 + 45° = 225° (3rd quadrant)

8 0
2 years ago
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