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3 years ago

How many triangles can be constructed with angles measuring 50º, 90º, and 40º?

Mathematics

2 answers:

NeX [460]3 years ago

6 0

Answer: Infinite

Step-by-step explanation:

We know that in a triangle the sum of all the interior angles must be 180°.

The given angles 50º, 90º and 40º

The sum of the angles 50º+ 90º + 40º= 180°

Thus, a triangle is possible with the given measurement.,

Let there is another triangle with the given angles, then by AAA similarity criteria they are similar.

Similarly, all the triangles with the same measurements of the angles must be similar.

Therefore, there are infinite number of triangles can be possible with angles measuring 50º, 90º, and 40º.

vodka [1.7K]3 years ago

6 0

If the angles of the triangle are 50, 90, 40 degrees, then only one triangle can be constructed, because if you change the sides the angles would change. <span />

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The length of a rectangle is five times its width.

Vera_Pavlovna [14]

Answer:

Step-by-step explanation:

l = 5w

A = l*w

180 = 5w *w

180 = 5w^2

Divide each side by 5

180/5 = 5w^2 /5

36 = w^2

Taking the square root of each side

sqrt(36) = sqrt(w)^2

6 =w

l = 5w = 5*6 = 30

The perimeter is

P = 2(l+w)

P = 2(6+30) = 2(36) = 72

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3 years ago

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Sunny_sXe [5.5K]

Answer:

Step-by-step explanation:

upper quartile is the median of the upper half of a data set

The median is the middle number which is 37

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3 years ago

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Solve for both systems: y+7z=22 and 8y + 7z = 127

7nadin3 [17]

Manipulate the first equation to
y = 22-7z

Then substitute in the second to find z
8(22-7z) + 7z = 127
176-56z+7z = 127
-49z = -49
z = 1

So y = 22-7(1) = 22-7 = 15

y = 15; z = 1

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3 years ago

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The number of hours worked per year per person in a state is normally distributed with a standard deviation of 39. A sample of 1

evablogger [386]

Answer:

The 98% confidence interval for the population mean number of hours worked per year per person is (2146, 2193).

Step-by-step explanation:

The question is incomplete.

The number of hours registered in the sample are:

2051 2061 2162 2167 2169 2171

2180 2183 2186 2195 2196 2198

2205 2210 2211

The sample mean can be calculated as:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{15}(2051+2061+2162+2167+2169+2171+2180+2183+2186+2195+2196+2198+2205+2210+2211)\\\\\\M=\dfrac{32545}{15}\\\\\\M=2169.67\\\\\\$