Find the value of x in the triangle shown below 4 3

Given the formula b=rs/t solve for s

Gekata [30.6K]

Answer:

$s = \frac{bt}{r}$

Step-by-step explanation:

To solve for s, you need to follow these steps.

First multiply by t both sides of the equation.

$b = \frac{rs}{t}$

$bt = \frac{rst}{t}$

We get $bt = rs$

Now we divide by r, both sides of the equation.

$\frac{bt}{r} = \frac{rs}{r}$

Finallly we obtain:

$\frac{bt}{r} = s$ or

$s = \frac{bt}{r}$

7 0

3 years ago

Define f(0,0) in a way that extends f to be continuous at the origin. f(x, y) = ln ( 19x^2 - x^2y^2 + 19 y^2/ x^2 + y^2) Let f (

kirill115 [55]

Answer:

f(0,0)=ln19

Step-by-step explanation:

$f(x,y)=ln(\frac{19x^2-x^2y^2+19y^2}{x^2+y^2})$ is given as continuous function, so there exist $lim_{(x,y)\rightarrow(0,0)}f(x,y)$ and it is equal to f(0,0).

Put x=rcosA annd y=rsinA

$f(r,A)=ln(\frac{19r^2cos^2A-r^2cos^2A*r^2sin^2A+19r^2sin^2A}{r^cos^2A+r^2sin^2A})=ln(\frac{19r^2(cos^2A+sin^2A)-r^4cos^2Asin^a}{r^2(cos^2A+sin^2A)})$

we know that $cos^2A+sin^2A=1$ , so we have that

$f(r,A))=ln(\frac{19r^2-r^4cos^2Asin^a}{r^2})=ln(19-r^2cos^2Asin^2A)$

$lim_{(x,y)\rightarrow(0,0)}f(x,y)=lim_{r\rightarrow0}f(r,A)=ln19$

So f(0,0)=ln19.

8 0

3 years ago

Please help thanks!!

Sidana [21]

So for this, you will be doing two different multiplications: 3 x 4 and √8 x √3.

3 x 4 = 12

√8 x √3 = √24

Now our result is 12√24, however, this can be simplified. Using the product rule of radicals (√ab = √a x √b), our simplification is such:

12√24 = 12√(8 x 3) = 12√(4 x 2 x 3) = 2 x 12√(2 x 3) = 24√6

In short, the answer is 24√6, or the first option.

4 0

2 years ago

One person wants to get a 95% z-confidence interval with a margin of error of at most 15 based on a population standard deviatio

Bess [88]

Answer:

$n=(\frac{1.96(60)}{15})^2 =61.46 \approx 62$

So the answer for this case would be n=62 rounded up to the nearest integer

Step-by-step explanation:

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".

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