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4 years ago

If u=1+3 i and v=-2-i what is u+v

Mathematics

2 answers:

Citrus2011 [14]4 years ago

6 0

The answer is 2-i
1+3=4
4-2=2
2-i=2-i

elena-14-01-66 [18.8K]4 years ago

4 0

<h2>Hello!</h2>

The answer is:

$u+v=-1+2i$

To solve the problem, we just need to consider that the variable "u" and "v" represents different expressions, and then, perform the operation and add the like terms.

We must remember that the like terms are the terms that share the same variable and the same exponent.

We have that:

$u=1+3i\\v=-2-i$

So, calculating we have:

$u+v=(1+3i)+(-2-i)=(1-2)+(3i-i)=-1+2i$

Hence, we have that:

$u+v=-1+2i$

Have a nice day!

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Find f(6) if f(x) = x^{2} ÷ 3 + x.

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6^2 is 36. / 3 makes 12. 12+6 is 18. final answer, most likely 18.

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3 years ago

Algebra tiles to prove that 4 + 8x and 4(2x+1)

Dafna1 [17]

Not sure what algebra tiles are

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I can prove it works at least most of the time by subsitute ing number for x

remember that 0,1,2 have special properties and are not very good for subsituting so
try 4
4+8(4)=4(2(4)+1)
true or false?
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multipily
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another way is to believe in the distributive property where
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7 0

3 years ago

Read 2 more answers

The realtor and her clients do not know the average home sale price for all of Guelph (500 was actually just a guess). However,

strojnjashka [21]

Answer:

a) The 99% confidence interval would be given by (346.708;453.292)

b) The 99% confidence interval would be given by (338.445;461.555)

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=400$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=80$ represent the population standard deviation

n=15 represent the sample size

2) Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$

Now we have everything in order to replace into formula (1):

$400-2.58\frac{80}{\sqrt{15}}=346.708$

$400+2.58\frac{80}{\sqrt{15}}=453.292$

So on this case the 99% confidence interval would be given by (346.708;453.292)

3) Part b

For this case we don't know the population deviation so we need to use the t distribution instead the normal standard distribution.

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

We need to find the degrees of freedom first df=n-1=15-1=14

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that $t_{\alpha/2}=2.98$