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3 years ago

How to find rationalizing factor of class 9

Mathematics

1 answer:

Mrrafil [7]3 years ago

8 0

Answer:

$\sqrt[5]{a^{3} b^{2}c}$

Step-by-step explanation:

Rationalization factor of $\sqrt[5]{a^2 b^3 c^4}$

$= \sqrt[5]{a^{5-2} b^{5-3}c^{5-4}}\\= \sqrt[5]{a^{3} b^{2}c^{1}}\\= \sqrt[5]{a^{3} b^{2}c}\\$

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fie abc un triunghi cu masura unghiului a de 90 de grade ab este 6 cm bc=10 si ca=8 calculati;sin b; tg b; sin c; tg c; cos b; c

lions [1.4K]

SinB = cosC = AC / BC = 8/10 = 4/5;
tgB = ctg C = AC/AB = 8/6 = 4/3, because we use T. Pitagora for calculating AB;
sinC = cosB = AB/ BC = 6/10 = 3/5;
tgC = ctgB = AB/ AC = 6/8 = 3/4.

6 0

3 years ago

PLEASE HELP. ILL PICK BRAINIEST ANSWER

klasskru [66]

The graph shows the height of a kicked soccer ball f(x), in feet, depending on the distance from the kicker x, in feet

Part A: What do the x-intercepts of the graph represent?

The x intercept means value of x when h(x)=0. That is, the x-intercept signify the distance from kicker when height h(x) =0. The x intercept is (0,0) and (16,0).

So, when the kicker is at a distance of 0 feet or 16 feet, height h(x)=0 feet.

What does maximum value of the graph represent?

The maximum value of the graph signify maximum height the soccer ball can achieve. The maximum height the graph has is at (8,10). That is, when kicker is at a distance of 8 feet from the ball, the ball attains a maximum height of 10 feet.

What are the intervals where the function is increasing and decreasing, and what do they represent about distance and height?

The increasing interval is (-∞,8]. That is, the height of the ball increases when kicker is increasing as the distance increases from -∞ to 8. But, the height keeps on decreasing from 8 to ∞. As, x, distance increases from 8, the height decreases, Hence, decreasing interval [8,∞)

Part B: What is an approximate average rate of change of the graph from x=8 to x=13, and what does this rate represent?

average rate of change= $\frac{f(x2)-f(x1)}{x2-x1}$

where x2=13 and x1=8. That is, the interval in which we are looking for average rate of change

f(x2)=f(13)=5

f(x1)=f(8)=10

average rate of change = $\frac{5-10}{13-8}$

= $\frac{-5}{5}$

=-1

average rate of change =-1.

This rate signify that height decreases by 1 feet as distance of the skipper increases by 1 feet (when distance, x is between 8 and 13)

7 0

3 years ago

Do one of the following, as appropriate. (a) Find the critical value z Subscript alpha divided by 2, (b) find the critical v

aliya0001 [1]

Answer:

$t=\pm 2.95$

Step-by-step explanation:

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".

The t distribution or Student’s t-distribution is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

Data given

Confidence =0.99 or 99%

$\alpha=1-0.99=0.01$ represent the significance level

n =16 represent the sample size

We don't know the population deviation $\sigma$

Solution for the problem

For this case since we don't know the population deviation and our sample size is <30 we can't use the normal distribution. We neeed to use on this case the t distribution, first we need to calculate the degrees of freedom given by:

$df=n-1=16-1=15$

We know that $\alpha=0.01$ so then $\alpha/2=0.005$ and we can find on the t distribution with 15 degrees of freedom a value that accumulates 0.005 of the area on the left tail. We can use the following excel code to find it:

"=T.INV(0.005;15)" and we got $t_{\alpha/2}=-2.95$ on this case since the distribution is symmetric we know that the other critical value is $t_{\alpha/2}=2.95$