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Strike441 [17]
4 years ago
5

Instructions: Determine if the two triangles in the image are congruent. If they are, state how you know by identifying the post

ulate.

Mathematics
1 answer:
icang [17]4 years ago
7 0

Answer: The triangles are congruent by HL (hypotenuse leg)

We have two right triangles. The tickmarks show the legs are congruent. The hypotenuses are congruent because they overlap perfectly. The hypotenuses are a shared common side. Using HL, we can prove the triangles to be congruent.

Side note: You could use the pythagorean theorem to find the measure of the third pair of sides, and that would lead to using SSS.

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Karla rode her bicycle 135 miles in 9 weeks, riding the same distance each week. Brad rode his bicycle 102 miles in 6 weeks, rid
blsea [12.9K]

Answer:

Karla's 15 miles per week to Brad's 17 miles per week.

Step-by-step explanation:

First, we would need find the total number of miles each individual rides per week. We do this by dividing the total miles ridden by the number of weeks it took for each individual like so...

Karla:  135 / 9 = 15 miles per week

Brad:   102 / 6 = 17 miles per week

Finally, the comparison would be

Karla's 15 miles per week to Brad's 17 miles per week.

8 0
3 years ago
Find the value of x. If your answer is not an integer express it in simplest radical form.
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5 0
4 years ago
Find the circumference of a circle with radius, r = 10.5m.Give your answer in terms of π<br> .
forsale [732]

Answer:

Circumference of circle =2πr=2π×10.5=21πm

Step-by-step explanation:

8 0
3 years ago
I need help to get good grade
8090 [49]

Answer:

The scale factor is now 3

Step-by-step explanation:

6/2=3

12/4=3

The smaller square was increased by the scale factor of 3

5 0
3 years ago
Find the roots of the equation<br> x ^ 2 + 3x-8 ^ -14 = 0 with three precision digits
scoray [572]

Answer:

Step-by-step explanation:

Given quadratic equation:

x^{2} + 3x - 8^{- 14} = 0

The solution of the given quadratic eqn is given by using Sri Dharacharya formula:

x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}

The above solution is for the quadratic equation of the form:

ax^{2} + bx + c = 0  

x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}

From the given eqn

a = 1

b = 3

c = - 8^{- 14}

Now, using the above values in the formula mentioned above:

x_{1, 1'} = \frac{- 3 \pm \sqrt{3^{2} - 4(1)(- 8^{- 14})}}{2(1)}

x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})})

x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})} - 3)

Now, Rationalizing the above eqn:

x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(- 8^{- 14})} - 3)\times (\frac{\sqrt{9 - 4(- 8^{- 14})} + 3}{\sqrt{9 - 4(- 8^{- 14})} + 3}

x_{1, 1'} = \frac{1}{2}.\frac{(\pm {9 - 4(- 8^{- 14})^{2}} - 3^{2})}{\sqrt{9 - 4(- 8^{- 14})} + 3}

Solving the above eqn:

x_{1, 1'} = \frac{2\times 8^{- 14}}{\sqrt{9 + 4\times 8^{-14}} + 3}

Solving with the help of caculator:

x_{1, 1'} = \frac{2\times 2.27\times 10^{- 14}}{\sqrt{9 + 42.27\times 10^{- 14}} + 3}

The precise value upto three decimal places comes out to be:

x_{1, 1'} = 0.758\times 10^{- 14}

5 0
3 years ago
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