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2 years ago

Pls help and show workings.

Mathematics

1 answer:

nekit [7.7K]2 years ago

3 0

Answer:

$\huge\boxed{\tt{SSS}}$

Step-by-step explanation:

SSS (Side-Side-Side) Postulate can be used to prove the two triangles congruent.

Whole figure is a parallelogram which means that opposite sides are parallel and equal (or congruent).

So, Two of the sides can be proved congruent by that way.

The third side is common in both of the triangles which means that the third side is also congruent.

Hence, the two triangles are congruent by SSS postulate.

$\rule[225]{225}{2}$

Hope this helped!

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Below are survival times (in days) of 13 guinea pigs that were injected with a bacterial infection in a medical study:

tresset_1 [31]

Outliers are data that are relatively far from other data elements.

The dataset has an outlier and the outlier is 120

The dataset is given as:

91 83 84 79 91 93 95 97 97 120 101 105 98

Sort the dataset in ascending order

79 83 84 91 91 93 95 97 97 98 101 105 120

<h3>The lower quartile (Q1)</h3>

The Q1 is then calculated as:

$Q1 = \frac{N +1}{4}th$

So, we have:

$Q1 = \frac{13 +1}{4}th$

$Q1 = \frac{14}{4}th$

$Q1 = 3.5th$

This is the average of the 3rd and the 4th element

$Q1 = \frac{1}{2} \times (84 + 91)$

$Q1 = 87.5$

<h3>The upper quartile (Q3)</h3>

The Q3 is then calculated as:

$Q3 = 3 \times \frac{N +1}{4}th$

So, we have:

$Q3 = 3 \times \frac{13 +1}{4}th$

$Q3 = 3 \times 3.5th$

$Q3 = 10.5th$

This is the average of the 10th and the 11th element.

$Q_3 =\frac12 \times (98 + 101)$

$Q_3 =99.5$

<h3>The interquartile range (IQR)</h3>

The IQR is then calculated as:

$IQR = Q_3 -Q_1$

$IQR = 99.5 - 87.5$

$IQR = 12$

Also, we have:

$IQR(1.5) = 12 \times 1.5$

$IQR(1.5) = 18$

<h3>The outlier range</h3>

The lower and the upper outlier range are calculated as follows:

$Lower = Q_1 - IQR(1.5)$

$Lower = 87.5- 18$

$Lower = 69.5$

$Upper = Q_3 + IQR(1.5)$

$Upper = 99.5 + 18$

$Upper = 117.5$

120 is greater than 117.5.

Hence, the dataset has an outlier and the outlier is 120

Read more about outliers at:

brainly.com/question/9933184

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2 years ago

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2 years ago

The combined area of two sqaures is 45 sqaure centimeters. Each side of one sqaure is twice as long as a side of the other sqaur

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Answer:the length of each side of the larger square is 6 centimeters.

Step-by-step explanation:

Let x represent the length of each side of the small square.

Let y represent the length of each side of the Large square.

The area of the small square is x^2

The area of the large square is y^2

The combined area of two squares is 45 square centimeters. This means that

x^2 + y^2 = 45 - - - - - - - - - - -1

Each side of one square is twice as long as a side of the other square. This means that

y = 2x

Substituting y = 2x into equation 1, it becomes

x^2 + (2x)^2 = 45

x^2 + 4x^2 = 45

5x^2 = 45

x^2 = 45/5 = 9

x = √9 = 3

Substituting x = 3 into y = 2x, it becomes

y = 2 × 3 = 6 centimeters

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2 years ago