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Greeley [361]
3 years ago
6

tyler claims that if two triangles each have a side length of 11 units and a side length of 8 units, and also an angle measuring

, they must be identical to each other. Do you agree? Explain your reasoning.

Mathematics
1 answer:
ExtremeBDS [4]3 years ago
3 0

Answer: Yes I agree, they must be identical to each other.

Step-by-step explanation: One of the conditions based on which triangles are similar is the SSA criteria, that is, Side-Side-Angle.

According to Tyler, two sides have been measured as 11 units and 8 units respectively in both triangles. If the corresponding angle is the same as indicated in the question, then the third side must be the same measurement. However, if the angles are of different sizes, then the third side would also be different for both triangles. For instance if the angle in triangle A is wider than that of triangle B, then the third side in triangle A would be longer than that of triangle B and in that case they would not be similar. Also if the angle in triangle A is narrower than that of triangle B, then the third side of triangle A would be shorter than that of triangle B and both triangles would not be similar as well.

Please refer to the picture attached. In the upper part of the picture, triangle A and triangle B are not similar though they both have sides measuring 11 and 8 units because angle x is not equal to angle y. However in the lower part, both triangles A and B are equal because both have angle x and sides 11 and 8 units, and that makes the third side have the same measurement.

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3 years ago
An automotive manufacturer wants to know the proportion of new car buyers who prefer foreign cars over domestic. Step 2 of 2 : S
ziro4ka [17]

Answer:

The 85% onfidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is (0.151, 0.205).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

Sample of 421 new car buyers, 75 preferred foreign cars. So n = 421, \pi = \frac{75}{421} = 0.178

85% confidence level

So \alpha = 0.15, z is the value of Z that has a pvalue of 1 - \frac{0.15}{2} = 0.925, so Z = 1.44.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.178 - 1.44\sqrt{\frac{0.178*0.822}{421}} = 0.151

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.178 + 1.44\sqrt{\frac{0.178*0.822}{421}} = 0.205

The 85% onfidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is (0.151, 0.205).

8 0
2 years ago
Joe is looking for a new laptop case. His computer is 18 x 13.5 inches. What is the perimeter and area of the laptop case?
sineoko [7]

Answer:

Perimeter =63 inches

Area = 243in²

Step-by-step explanation:

Given data

Length of computer = 18in

Width of computer = 13.5in

From the given data the computer has a rectangular shape

Perimeter of the computer = L+L+W+W= 2L+2W

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3 0
3 years ago
Read 2 more answers
Y''+y'+y=0, y(0)=1, y'(0)=0
mars1129 [50]

Answer:

y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form ay''+by'+cy=0.

Given: y''+y'+y=0

Let y=e^{rt} be it's solution.

We get,

\left ( r^2+r+1 \right )e^{rt}=0

Since e^{rt}\neq 0, r^2+r+1=0

{ we know that for equation ax^2+bx+c=0, roots are of form x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} }

We get,

y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}

For two complex roots r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta, the general solution is of form y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )

i.e y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

Applying conditions y(0)=1 on e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right ), c_1=1

So, equation becomes y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

On differentiating with respect to t, we get

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Applying condition: y'(0)=0, we get 0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}

Therefore,

y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

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3 years ago
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Your ans is 1 
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