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Elden [556K]
3 years ago
5

David weeds lawns in his neighborhood to earn money to buy video games. For each lawn, he charges $5.50. If he weeds 4 lawns on

Saturday, how much money will he earn?
Mathematics
2 answers:
user100 [1]3 years ago
6 0

Answer:

$22

Step-by-step explanation:

4 * 5.50 = 22

yawa3891 [41]3 years ago
4 0

Answer:

$51.75

Step-by-step explanation:

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3 years ago
Write the sentence as an equation.<br> The sum of 24 and x is - 20.
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3 years ago
Read 2 more answers
In this triangle, which of the following is true?​
Svetradugi [14.3K]

We know that this is a right triangle, so one of the sides has to be 90 (indicated by the square on the corner)  The angles of all triangles must add to exactly 180.  Subtract 90 and 35 from 180

If we do that, we get 55.  We know that the unknown angle is 55 now.  Now, it asks us for b.  We can pythagoras theorem which states that (a^2 + b^2 = c^2)  We already know c (c is 20)  It does not tell us 2 of the sides but we do know that the answer is either B or D.  

The side opposite of 90* should be the largest.  The size opposite of 55 would be the second largest one.  The side opposite of 35 would be the smallest one.  Using pythagoras theorem again, I can try plugging in both 11.47 and 16.38 with c^2 - b^2 = a^2  

11.47:

400 - 132 =268  (sq root now = 16.37) this would mean that A is larger than B.  B should be larger than A.  Try the other one

16.37:

400 -268 = 132 (sq root = 16.37) this would mean that B is larger than A which is what we want.

Thus, 16.37  = B and the last option would be correct (D)

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7 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
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