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OleMash [197]
3 years ago
12

I am 67 less than 85. What number am I?

Mathematics
2 answers:
astra-53 [7]3 years ago
7 0
I think the answer is 22
Luba_88 [7]3 years ago
5 0
The answer is 18 can you help me with mine
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3a+b=225, b-a=25 solve.
adelina 88 [10]
If you are solving by substituting A=50 and B=75
5 0
4 years ago
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Consider the following series. 1 4 + 1 8 + 1 12 + 1 16 + 1 20 Determine whether the series is convergent or divergent. Justify y
patriot [66]

Answer:

The series diverges

Step-by-step explanation:

First, let's find the sequence:

4, 8, 12, 16, 20,...,

As we can see, they are multiples of 4, so one possible sequence is:

4,8,12,16,20,...,=4n\hspace{8};\hspace{3}n\in Z

Hence, we can represent the series as:

$$\sum_{n=1}^{\infty} 4n

Let's use the zero test which states:

$$\Sigma a_n$$\\Diverges\hspace{3}if\\\\ \lim_{n \to \infty} a_n \neq 0

So:

\lim_{n \to \infty} 4n= 4 \lim_{n \to \infty} n\\ \\ \lim_{n \to \infty} n=\infty\\ \\4\infty= \infty\\\\Hence\\\\ \lim_{n \to \infty} 4n=\infty

Therefore, by the zero test, the series diverges

3 0
3 years ago
I need help, in 7-12. Please, someone??
ratelena [41]
20/29
16/34
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5 0
3 years ago
Prove that if $w,z$ are complex numbers such that $|w|=|z|=1$ and $wz\ne -1$, then $\frac{w+z}{1+wz}$ is a real number.
mr_godi [17]

Answer:

See proof below

Step-by-step explanation:

Let r=\frac{w+z}{1+wz}. If w=-z, then r=0 and r is real. Suppose that w≠-z, that is, r≠0.

Remember this useful identity: if x is a complex number then x\bar{x}=|x|^2 where \bar{x} is the conjugate of x.

Now, using the properties of the conjugate (the conjugate of the sum(product) of two numbers is the sum(product) of the conjugates):

\frac{r}{\bar{r}}=\frac{w+z}{1+wz} \left(\frac{1+\bar{w}\bar{z}}{\bar{w}+\bar{z}}{\right)

=\frac{(w+z)(1+\bar{w}\bar{z})}{(1+wz)(\bar{w}+\bar{z})}=\frac{w+z+w\bar{w}\bar{z}+z\bar{z}\bar{w}}{\bar{w}+\bar{z}+\bar{w}wz+\bar{z}zw}=\frac{w+z+w+|w|^2\bar{z}+|z|^2\bar{w}}{\bar{w}+\bar{z}+|w|^2z+|z|^2w}=\frac{w+z+\bar{z}+\bar{w}}{\bar{w}+\bar{z}+z+w}=1

Thus \frac{r}{\bar{r}}=1. From this, r=\bar{r}. A complex number is real if and only if it is equal to its conjugate, therefore r is real.

3 0
3 years ago
To the nearest tenth of a precent, 78 is what precent of 82
forsale [732]

Answer:

95.1

Step-By-Step Explanation

78:82*100 =

(78*100):82 =

7800:82 = 95.12

Now we have: 78 is what percent of 82 = 95.12

Question: 78 is what percent of 82?

Percentage solution with steps:

Step 1: We make the assumption that 82 is 100% since it is our output value.

Step 2: We next represent the value we seek with x.

Step 3: From step 1, it follows that 100%=82

Step 4: In the same vein, x%=78

Step 5: This gives us a pair of simple equations:

$100\%=82(1)

$x\%=78(2)

Step 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS

(left hand side) of both equations have the same unit (%); we have

100% over x% = 82 over 78

Step 7: Taking the inverse (or reciprocal) of both sides yield

x% over 100% = 78 over 82

x = 95.12%

Therefore, 78 is 95.12% of 82

I only 1 more Brainliest to get to the next level so getting one would be lovely

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3 years ago
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