All categories

2 years ago

Which dimensions can create only one unique triangle?

Mathematics

1 answer:

s2008m [1.1K]2 years ago

3 0

Answer:

<h2>The dimensions must all add up to 180°.</h2>

Step-by-step explanation:

For example;

Your triangles' angles are: 56, 77, and 47.

56 + 77 + 47 = 180

*You did not give me any values to solve for, but just since you didn't- I'v provided an example to assist you in doing it yourself. If you want me to solve your full equation please add the angles or 3 dimensions of the triangle.

<h3><em>I hope this helped!</em></h3>

You might be interested in

Yuuma applied the steps below to find the product of (–10.2)(–16.4).

goblinko [34]

Answer:

164+3.28

Step-by-step explanation:

(16.4)(10)+(16.4)(0.2)

164+3.28

167.28

4 0

1 year ago

Helppppppppopopoppppp

skelet666 [1.2K]

Answer:

DG = 30

Step-by-step explanation:

Given:

DH = 6

DE = 4

EF = 16

Required:

Solution:

DG = DH + HG

DG = 6 + HG

Let's find HG

Given that HE is parallel to the third side of ∆DGF, based on the side-splitter theorem, the other two sides of ∆DGF are divided proportionally.

Therefore,

DH/HG = DE/EF

6/HG = 4/16

Cross multiply

HG*4 = 16*6

HG = 96/4

HG = 24

✔️DG = 6 + HG

DG = 6 + 24

DG = 30

7 0

2 years ago

-6 + x over 4=-5 what is the value of x

My name is Ann [436]

Answer:x=26

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78

____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

$m-t_{a/2,n-1}\frac{s}{\sqrt{n} }$ ≤ μ ≤ $m+t_{a/2,n-1}\frac{s}{\sqrt{n} }$

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and $t_{a/2,n-1}$ is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and $t_{0.05,9}$ by 1.8331, we get that the 90% confidence interval for the mean speed is: