HELP

Find the area of the figure will give brain illest if correct!

Travka [436]

I’m sure that the area is 80. If you do 10 times 5, that will be 50.Then if you do 6 times 5, for the upper remaining part of the figure it would be 30. So 30 plus 50 is 50. Hope this helps!

3 0

3 years ago

If f(x)=x-6 and g(x)=x1/2(x+3), find g(x) times f(x)

Elden [556K]

To solve this problem you must apply the proccedure shown below:
1. You have the following functions given in the problem above:
$f(x)=x-6 
$ and $g(x)=x1/2(x+3)=x(x+3)/2$ 
2. You have that g(x) times f(x) can be written as g(x)*f(x). Therefore, you must multiply f(x) * g(x), as following:
 $g(x)*f(x)=(x-6)*x(x+3)/2$ 
3. Applying the distributive property, you have:
 $g(x)*f(x)=( x^{3} + 3x^2 - 6x^{2} -18x)/2$
$g(x)*f(x)=(x^{3} - 3x^{2} -18x)/2$
The answer is: $g(x)*f(x)=(x^{3} - 3x^{2} -18x)/2$

5 0

4 years ago

In a dance competition, a participant has to score a total of at least 30 points in the first four rounds combined to move on to

77julia77 [94]

So Steward scored 5 for the first round, that will be our constant. There is 3 more rounds for him to score at least 30. at least tells us this is a greater than or equal to 5+3p<30

3 0

3 years ago

The interior angles of a hexagon are in the ratio 1:2:3:4:5:9 Find the angles. This is an example of a concave hexagon. Write in

algol [13]

Answer:hope this helps.. please mark as brainliest.....

8 0

3 years ago

Read 2 more answers

The score on an exam from a certain MAT 112 class, X, is normally distributed with μ=78.1 and σ=10.8.

salantis [7]

a) $X$

b) 0.1539

c) 0.1539

d) 0.6922

Step-by-step explanation:

a)

In this problem, the score on the exam is normally distributed with the following parameters:

$\mu=78.1$ (mean)

$\sigma = 10.8$ (standard deviation)

We call X the name of the variable (the score obtained in the exam).

Therefore, the event "a student obtains a score less than 67.1) means that the variable X has a value less than 67.1. Mathematically, this means that we are asking for:

$X$

And the probability for this to occur can be written as:

$p(X$

b)

To find the probability of X to be less than 67.1, we have to calculate the area under the standardized normal distribution (so, with mean 0 and standard deviation 1) between $z=-\infty$ and $z=Z$ , where Z is the z-score corresponding to X = 67.1 on the s tandardized normal distribution.

The z-score corresponding to 67.1 is:

$Z=\frac{67.1-\mu}{\sigma}=\frac{67.1-78.1}{10.8}=-1.02$

Therefore, the probability that X < 67.1 is equal to the probability that z < -1.02 on the standardized normal distribution:

$p(X$

And by looking at the z-score tables, we find that this probability is:

$p(z$

And so,

$p(X$

c)

Here we want to find the probability that a randomly chosen score is greater than 89.1, so

$p(X>89.1)$

First of all, we have to calculate the z-score corresponding to this value of X, which is:

$Z=\frac{89.1-\mu}{\sigma}=\frac{89.1-78.1}{10.8}=1.02$

Then we notice that the z-score tables give only the area on the left of the values on the left of the mean (0), so we have to use the following symmetry property:

$p(z>1.02) =p(z$