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In-s [12.5K]
3 years ago
12

1.) (2)(8+2i)= Multiplying complex numbers

Mathematics
2 answers:
Anna35 [415]3 years ago
7 0

Answer:

Step-by-step explanation:

7+3i

Nutka1998 [239]3 years ago
4 0
Your answer is 16+4i
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a scale drawing of a building needs to be made using the scale 1 in = 190 ft. how tall will the builging in the scale drawing be
Artist 52 [7]

Answer:

\boxed{\text{3 in}}

Step-by-step explanation:

1 in = 190 ft

x in = 570 ft

We can use ratio and proportion to solve this problem.

\begin{array}{cccl}\dfrac{1}{x} & = & \dfrac{190}{570} & \\\\570 & = & 190x & \text{Mutiplied by lowest common denominator (570x)}\\3 & = & x & \text{Divided each side by 3}\\\end{array}

The building will be \boxed{\textbf{3 in}} tall in the scale drawing.

3 0
3 years ago
PLEASE HELP ME IM STRUGGLING!!!
Kitty [74]

Answer:

The required answer is c=7\sqrt{3}

Therefore the number in green box should be 7.

Step-by-step explanation:

Given:

AB = 7√2

AD = a , BD = b , DC = c , AC = d

∠B = 45°, ∠C = 30°

To Find:

c = ?

Solution:

In Right Angle Triangle ABD Sine identity we have

\sin B = \dfrac{\textrm{side opposite to angle B}}{Hypotenuse}\\

Substituting the values we get

\sin 45 = \dfrac{AD}{AB}= \dfrac{a}{7\sqrt{2}}

\dfrac{1}{\sqrt{2}}= \dfrac{a}{7\sqrt{2}}\\\\\therefore a=7

Now in Triangle ADC Tangent identity we have

\tan C = \dfrac{\textrm{side opposite to angle C}}{\textrm{side adjacent to angle C}}

Substituting the values we get

\tan 30 = \dfrac{AD}{DC}= \dfrac{a}{c}\\\\\dfrac{1}{\sqrt{3}}=\dfrac{7}{c}\\\\\therefore c=7\sqrt{3}

The required answer is c=7\sqrt{3}

8 0
3 years ago
Edward walks at a pace 2 1/4 miles in 2/3 hour. How many miles does Edward walk per hour
GuDViN [60]

Divide distance walked by time:

2 1/4 miles / 2/3 hours = 3 3/8 miles per hour

8 0
3 years ago
90 km is 8% of what distance
sergejj [24]
1,125 km is the distance
4 0
3 years ago
Please solve! Evaluate the function.
nikitadnepr [17]

Answer:

(h o g)(5) = 20

Step-by-step explanation:

(h o g)(5) = h(g(5)) <em>This is two ways of writing the expression</em>

g(x) = \sqrt{5x}  h(x) = 3x + 5 <em>g(x) and h(x) are your "plug-ins" for the expression</em>

g(5) = \sqrt{5(5)} = \sqrt{25} = 5 <em>This is how you would solve for plugging in 5 to g(x)</em>

<em>Whatever you get for g(5) {the answer to it} is what you will be plugging into h(x); x being equal to g(5).</em>

h(5) = 3(5) + 5 = 15 + 5 = 20  

6 0
3 years ago
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