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Marysya12 [62]
3 years ago
11

Find the area of the regular polygon.

Mathematics
1 answer:
blondinia [14]3 years ago
8 0

Since the polygon is a square, the diagonal is \sqrt{2} times the side of the square. The diagonal is 14 yards, so the side of the square is 7\sqrt{2} yards. Since the area of a square is it's side squared, the area is (7\sqrt{2})^{2}, which is 98 square yards.

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A quadrilateral is graphed in the coordinate plane below. Which classification best describes the quadrilateral (parallelogram,
Ahat [919]

Answer:

Trapezoid

Step-by-step explanation:

Given quadrilateral has vertices at points A(-2,-1), B(3,13), C(15,5) and D(13,-11).

Find slopes of lines AD and BC:

\text{Slope}_{AD}=\dfrac{y_D-y_A}{x_D-x_A}=\dfrac{-11-(-1)}{13-(-2)}=\dfrac{-11+1}{13+2}=\dfrac{-10}{15}=-\dfrac{2}{3}\\ \\\text{Slope}_{BC}=\dfrac{y_C-y_B}{x_C-x_B}=\dfrac{5-13}{15-3}=\dfrac{-8}{12}=-\dfrac{2}{3}

Since the slopes are the same, lines AD and BC are parallel.

Find slopes of lines ABD and CD:

\text{Slope}_{AB}=\dfrac{y_B-y_A}{x_B-x_A}=\dfrac{13-(-1)}{3-(-2)}=\dfrac{14}{5}\\ \\\text{Slope}_{CD}=\dfrac{y_D-y_C}{x_D-x_C}=\dfrac{-11-5}{13-15}=\dfrac{-16}{-2}=8

Since the slopes are different, lines AB and CD are not parallel.

This means quadrilateral ABCD is trapezoid (two opposite sides - parallel and two another opposite sides - not parallel)

7 0
4 years ago
3(n+5)≥3n+8. help, please
SCORPION-xisa [38]

Answer:

so you distrubute the 3 into the parenthesis and you then get 3n+15 is greater than or equal to 3n+8 then you subtract the 3n from the right side from the 3n on the left side and you are left with 15 is greater than or equal to 8

Step-by-step explanation:

4 0
3 years ago
If q is between 0 and 90 and tan(theta)= 7/8, find costheta
Leya [2.2K]

Answer:

hope it helps you see the attachment for further information.....

8 0
3 years ago
Please Help ! Will Reward Brainliest
pychu [463]

I think 42

would be the best answer to your question in my opininon.


7 0
4 years ago
The CEO of the Jen and Benny's ice cream company is concerned about the net weight of ice cream in their 50 ounce ice cream tubs
MissTica

Answer:

t=\frac{54.98-52}{\frac{8.43}{\sqrt{26}}}=1.8025  

We need to find the degrees of freedom given by:

df = n-1= 26-1 = 25

Since is a right tailed test the p value would be:  

p_v =P(t_{25}>1.8025)=0.0418  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average is higher than 52 at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=54.98 represent the sample mean

s=8.43 represent the sample deviation

n=26 sample size  

\mu_o =52 represent the value that we want to test  

\alpha=0.05 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is higher than 52, the system of hypothesis would be:  

Null hypothesis:\mu \leq 52  

Alternative hypothesis:\mu > 52  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{54.98-52}{\frac{8.43}{\sqrt{26}}}=1.8025  

P-value  

We need to find the degrees of freedom given by:

df = n-1= 26-1 = 25

Since is a right tailed test the p value would be:  

p_v =P(t_{25}>1.8025)=0.0418  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the average is higher than 52 at 5% of signficance.  

7 0
3 years ago
Read 2 more answers
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