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3 years ago

Prove how they are congruent. Geometry:SSS

Mathematics

2 answers:

Andrej [43]3 years ago

5 0

The additional information needed to prove that both triangles are congruent by the SSS Congruence Theorem would be: C. HJ ≅ LN

Recall:

Based on the Side-Side-Side Congruence Theorem, (SSS), two triangles can be said to be congruent to each other if they have three pairs of congruent sides.

Thus, in the two triangles given, the two triangles has:

Two pairs of congruent sides - HI ≅ ML and IJ ≅ MN

Therefore, an additional information needed to prove that both triangles are congruent by the SSS Congruence Theorem would be: C. HJ ≅ LN

Learn more about SSS Congruence Theorem on:

brainly.com/question/4280133

Stels [109]3 years ago

4 0

$\huge \bf༆ Answer ༄$

For the given triangles to be congruent by SSS criterion, the sides HJ and LN should be equal ~

therefore correct choice is ~ C

$\large \boxed{ \sf \: HJ \cong LN}$

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3 years ago

While conducting experiments, a marine biologist selects water depths from a uniformly distributed collection that vary between

aleksandr82 [10.1K]

Answer:

The probability that a randomly selected depth is between 2.25 m and 5.00 m is 0.55.

Step-by-step explanation:

Let the random variable X denote the water depths.

As the variable water depths is continuous variable, the random variable X follows a continuous Uniform distribution with parameters a = 2.00 m and b = 7.00 m.

The probability density function of X is:

$f_{X}(x)=\frac{1}{b-a};\ a$

Compute the probability that a randomly selected depth is between 2.25 m and 5.00 m as follows:

$P(2.25$

$=\frac{1}{5.00}\int\limits^{5.00}_{2.25} {1} \, dx\\\\=0.20\times [x]^{5.00}_{2.25} \\\\=0.20\times (5.00-2.25)\\\\=0.55$

Thus, the probability that a randomly selected depth is between 2.25 m and 5.00 m is 0.55.

6 0

3 years ago

The temperature fell from 0 Degrees Fahrenheit to 15 and one-half Degrees Fahrenheit below 0 in 5 and three-fourths hours. Wen t

Kryger [21]

Answer:

The correct answer will be:

$-\dfrac{62}{23}$

Step-by-step explanation:

It is given that :

Initial temperature, $T_1 = 0^\circ F$

Final temperature,

$T_2 = -15\dfrac{1}{2}^\circ F\\\Rightarrow T_2 = -\dfrac{15\times 2+1}{2} ^\circ F\\\Rightarrow T_2 = -\dfrac{31}{2} ^\circ F$

Time taken :

$5\dfrac{3}{4}\ hrs = \dfrac{5 \times 4+3}{4}\ hrs = \dfrac{23}{4}\ hrs$

Change in temperature per hour:

$\dfrac{\text{Difference of temperature}}{\text{Total Time Taken}}\\\Rightarrow \dfrac{T_2-T_1}{\text{Total Time Taken}}$

Putting the values of temperatures and time:

$\dfrac{\dfrac{-31}{2}-0}{\dfrac{23}{4}}\\\Rightarrow \dfrac{\dfrac{-31}{2}}{\dfrac{23}{4}}\\\Rightarrow \dfrac{-31 \times 4}{2 \times 23}} \text{---- Error done by Wen at this step}\\\Rightarrow \dfrac{-31 \times 2}{23}}\\\Rightarrow \dfrac{-62}{23}}$