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4 years ago

Why does the LL theorem hold for proving right triangles congruent?

Mathematics

2 answers:

Rufina [12.5K]4 years ago

5 0

B.) The right angle is included between the legs.

alina1380 [7]4 years ago

3 0

Answer:

(A) The legs in right triangles are equal.

Step-by-step explanation:

LL theorem or Leg-Leg theorem: The LL theorem is also known as the leg-leg theorem. It states that if the legs of one right triangle are congruent to the legs of another right triangle, then the triangles are congruent.

Thus, in order to prove that the two right triangles are congruent,w use the LL theorem which holds that The legs in right triangles are equal.

Hence, option A is correct.

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The perimeter of a rectangle is 34 inches. If the length of the rectangle is 11 inches, which equation could be used to find the

KATRIN_1 [288]

Answer:

長方形の周囲は34インチです。長方形の長さが11インチの場合、どの式を使用して幅を見つけることができますか？バツ？

あなたの質問に答えることはできませんが、あなたはこれを翻訳するのがとても賢いです。

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3 years ago

Eloy went shopping and bought 2 shirts for $12.95 each, a pair of pants for $35.89, and a hat for $11.47. Eloy paid with a $100

grin007 [14]

Answer:

26.74

Step-by-step explanation:

(12.95x2)+35.89+11.47=73.26

100-73.26=26.74

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3 years ago

How is finding the unit rate helpful before graphing a proportional relationship

eduard

It is helpful to find a unit rate first. If you find a unit rate first, then it will be easier to compare

6 0

3 years ago

Read 2 more answers

(d). Use an appropriate technique to find the derivative of the following functions:

natima [27]

(i) I would first suggest writing this function as a product of the functions,

$\displaystyle y = fgh = (4+3x^2)^{1/2} (x^2+1)^{-1/3} \pi^x$

then apply the product rule. Hopefully it's clear which function each of f, g, and h refer to.

We then have, using the power and chain rules,

$\displaystyle \frac{df}{dx} = \frac12 (4+3x^2)^{-1/2} \cdot 6x = \frac{3x}{(4+3x^2)^{1/2}}$

$\displaystyle \frac{dg}{dx} = -\frac13 (x^2+1)^{-4/3} \cdot 2x = -\frac{2x}{3(x^2+1)^{4/3}}$

For the third function, we first rewrite in terms of the logarithmic and the exponential functions,

$h = \pi^x = e^{\ln(\pi^x)} = e^{\ln(\pi)x}$

Then by the chain rule,

$\displaystyle \frac{dh}{dx} = e^{\ln(\pi)x} \cdot \ln(\pi) = \ln(\pi) \pi^x$

By the product rule, we have

$\displaystyle \frac{dy}{dx} = \frac{df}{dx}gh + f\frac{dg}{dx}h + fg\frac{dh}{dx}$

$\displaystyle \frac{dy}{dx} = \frac{3x}{(4+3x^2)^{1/2}} (x^2+1)^{-1/3} \pi^x - (4+3x^2)^{1/2} \frac{2x}{3(x^2+1)^{4/3}} \pi^x + (4+3x^2)^{1/2} (x^2+1)^{-1/3} \ln(\pi) \pi^x$

$\displaystyle \frac{dy}{dx} = \frac{3x}{(4+3x^2)^{1/2}} \frac{1}{(x^2+1)^{1/3}} \pi^x - (4+3x^2)^{1/2} \frac{2x}{3(x^2+1)^{4/3}} \pi^x + (4+3x^2)^{1/2} \frac{1}{ (x^2+1)^{1/3}} \ln(\pi) \pi^x$

$\displaystyle \frac{dy}{dx} = \boxed{\frac{\pi^x}{(4+3x^2)^{1/2} (x^2+1)^{1/3}} \left( 3x - \frac{2x(4+3x^2)}{3(x^2+1)} + (4+3x^2)\ln(\pi)\right)}$

You could simplify this further if you like.

In Mathematica, you can confirm this by running

D[(4+3x^2)^(1/2) (x^2+1)^(-1/3) Pi^x, x]

The immediate result likely won't match up with what we found earlier, so you could try getting a result that more closely resembles it by following up with Simplify or FullSimplify, as in

FullSimplify[%]

(% refers to the last output)

If it still doesn't match, you can try running

Reduce[<our result> == %, {}]

and if our answer is indeed correct, this will return True. (I don't have access to M at the moment, so I can't check for myself.)

(ii) Given

$\displaystyle \frac{xy^3}{1+\sec(y)} = e^{xy}$

differentiating both sides with respect to x by the quotient and chain rules, taking y = y(x), gives

$\displaystyle \frac{(1+\sec(y))\left(y^3+3xy^2 \frac{dy}{dx}\right) - xy^3\sec(y)\tan(y) \frac{dy}{dx}}{(1+\sec(y))^2} = e^{xy} \left(y + x\frac{dy}{dx}\right)$

$\displaystyle \frac{y^3(1+\sec(y)) + 3xy^2(1+\sec(y)) \frac{dy}{dx} - xy^3\sec(y)\tan(y) \frac{dy}{dx}}{(1+\sec(y))^2} = ye^{xy} + xe^{xy}\frac{dy}{dx}$

$\displaystyle \frac{y^3}{1+\sec(y)} + \frac{3xy^2}{1+\sec(y)} \frac{dy}{dx} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} \frac{dy}{dx} = ye^{xy} + xe^{xy}\frac{dy}{dx}$

$\displaystyle \left(\frac{3xy^2}{1+\sec(y)} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} - xe^{xy}\right) \frac{dy}{dx}= ye^{xy} - \frac{y^3}{1+\sec(y)}$

$\displaystyle \frac{dy}{dx}= \frac{ye^{xy} - \frac{y^3}{1+\sec(y)}}{\frac{3xy^2}{1+\sec(y)} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} - xe^{xy}}$

which could be simplified further if you wish.

In M, off the top of my head I would suggest verifying this solution by

Solve[D[x*y[x]^3/(1 + Sec[y[x]]) == E^(x*y[x]), x], y'[x]]

but I'm not entirely sure that will work. If you're using version 12 or older (you can check by running $Version), you can use a ResourceFunction,

ResourceFunction["ImplicitD"][<our equation>, x]

but I'm not completely confident that I have the right syntax, so you might want to consult the documentation.

3 0

2 years ago

I WILL GIVE U BRAINLIEST AND THANKS AND 5 STARS

Effectus [21]

Answer:

I agree with Noah's Method because following PEMDAS you have to do exponents first. 10^3 is 1000 and then you can add the 20. Jada's method is incorrect because you cannot add the two together.

6 0

3 years ago