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Ber [7]
4 years ago
9

What are the values of x and y if this equation is true? 22(x + yi) + (28 + 4i) 72 – 62i

Mathematics
1 answer:
ASHA 777 [7]4 years ago
8 0

Answer:

x = 2

y = -3

Step-by-step explanation:

The given equation is,

22(x + yi) + (28 + 4i) = 72 - 62i

By solving this equation further,

22x + 22yi + 28 + 4i = 72 - 62i

(22x + 28) + (22y + 4)i = 72 - 62i

Now both the sides of the equation is in the form of complex number,

By comparing real and imaginary parts given on both the sides,

22x + 28 = 72

22x = 72 - 28

22x = 44

x = 2

22y + 4 = -62

22y = -62 - 4

22y = -66

y = -3

Therefore, x = 2 and y = -3 are the values for which the given equation is true.

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the hu family goes out for lunch, and the price of the meal is $43. The sales tax on the meal is 6%, and the family also leaves
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Answer:

Step-by-step explanation:

43.00x.06=2.58

43.00x.15=6.45

43.00+2.58+6.45=54.03

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Answer:

5

Step-by-step explanation:

The clothing store already sold $300 .

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They’re $200 short of their goal.

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Find the tenth term of the geometric sequence, given the first term and common ratio.
Natalija [7]

Answer:

a_{10}=\dfrac{1}{128}

Step-by-step explanation:

In the geometric series:

a_1=4\\ \\r=\dfrac{1}{2}

The nth term of the geometric sequence can be calculated using formula

a_n=a_1\cdot r^{n-1}

In your case, n = 10, then

a_{10}\\ \\=4\cdot \left(\dfrac{1}{2}\right)^{10-1}\\ \\=4\cdot \left(\dfrac{1}{2}\right)^9\\ \\=2^2\cdot \dfrac{1}{2^9}\\ \\=\dfrac{1}{2^{9-2}}\\ \\=\dfrac{1}{2^7}\\ \\=\dfrac{1}{128}

8 0
3 years ago
Help! If you know this can you tell me how to do it?
aleksandr82 [10.1K]

Answer:

c

Step-by-step explanation:

Here's how this works:

Get everything together into one fraction by finding the LCD and doing the math.  The LCD is sin(x) cos(x).  Multiplying that in to each term looks like this:

[sin(x)cos(x)]\frac{sin(x)}{cos(x)}+[sin(x)cos(x)]\frac{cos(x)}{sin(x)} =?

In the first term, the cos(x)'s cancel out, and in the second term the sin(x)'s cancel out, leaving:

\frac{sin^2(x)}{sin(x)cos(x)}+\frac{cos^2(x)}{sin(x)cos(x)}=?

Put everything over the common denominator now:

\frac{sin^2(x)+cos^2(x)}{sin(x)cos(x)}=?

Since sin^2(x)+cos^2(x)=1, we will make that substitution:

\frac{1}{sin(x)cos(x)}

We could separate that fraction into 2:

\frac{1}{sin(x)}×\frac{1}{cos(x)}

\frac{1}{sin(x)}=csc(x)  and  \frac{1}{cos(x)}=sec(x)

Therefore, the simplification is

sec(x)csc(x)

5 0
4 years ago
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