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Monica [59]
2 years ago
6

One of the legs of a right triangle measures 15 cm and its hypotenuse measures 17 cm. Find the measure of the other leg. If nece

ssary, round to the nearest tenth.
Mathematics
2 answers:
Sergeeva-Olga [200]2 years ago
8 0

Answer:

8

Step-by-step explanation:

The Pythagorean theorem shows us that the measure of the sides in a right triangle can be calculated using the formula a^2 + b^2 = c^2. Using this formula, and filling in the blank, we can conclude that the other leg measures 8 cm.

Hitman42 [59]2 years ago
3 0

Answer:

15

Step-by-step explanation:

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Hi quick whats 18 x 1.25 <br> step by step !!
Ivan

Answer:

22.5

Step-by-step explanation:

125

x 18

---------

1. first set up your problem

2 4

125

x 18

------

1000

2. next multiply the 8 by 5 getting 40, drop the 0 anf move the 4 to the top above 2.

3. multiply 8 by 2 getting 16, then add the 4 you have ontop of the 2 to the 16 getting 20. Drop the 0 and bring the 2 over and place it above the one.

4. multiply 8 by 1 getting 8 then add the 2 above the 1 to 8 getting 10. Then put the ten infront of the tw zeros. 1000

125

x 18

--------

1000

0

Next add a 0 as a place filler under the 0 at the end.

125

x 18

-------

1000

1250

Then multiply the 1 by 5, then 1 by 2, then finnaly 1 by 1, getting you 1250.

1000

+1250

add together

getting 2250

then move 2 decimal places over to get 22.5

3 0
3 years ago
Please Help, this is due by 4
lys-0071 [83]
B. (4, 4) is a solution to the equation.
4 0
3 years ago
Plz help me to answer
IgorLugansk [536]

22/3 sorry I accidentally deleted steps

8 0
2 years ago
Find dy/dx of the function y = √x sec*-1 (√x)​
ELEN [110]

Hi there!

\large\boxed{\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) +  \frac{1}{2|\sqrt{x}|\sqrt{{x} - 1}}}

y = \sqrt{x} * sec^{-1}(-\sqrt{x}})

Use the chain rule and multiplication rules to solve:

g(x) * f(x) = f'(x)g(x) + g'(x)f(x)

g(f(x)) = g'(f(x)) * 'f(x))

Thus:

f(x) = √x

g(x) = sec⁻¹ (√x)

\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \sqrt{x} * \frac{1}{\sqrt{x}\sqrt{\sqrt{x}^{2} - 1}} * \frac{1}{2\sqrt{x}}

Simplify:

\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \sqrt{x} * \frac{1}{2|x|\sqrt{{x} - 1}}

\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) +  \frac{1}{2|\sqrt{x}|\sqrt{{x} - 1}}

5 0
3 years ago
Read 2 more answers
a 50 ft pole has a support wire that runs from its top to the ground with an angle of depression of 75°. How far from the base o
lukranit [14]
The height of the pole is the vertical leg of the right triangle. The angle of depression is an exterior angle to the triangle at its upper angle. The angle of depression is complementary to the interior angle, so the measure of the interior angle is 15. The base angles are 90 and 75. Use the tangent ratio now, using the base angle of 75 and the height of the pole being 50: tan75=50/x.
Make sure your calculator is in degree mode to do this and get that x = 13.4
4 0
3 years ago
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