If the given is DEFG is a kite how do I prove that triangle EFH is congruent to triangle GFH

Mathematics

1 answer:

nordsb [41]2 years ago

4 0

Answer:

They both have H hoel

Step-by-step explanation:

You might be interested in

What is the value of x

aivan3 [116]

Answer:

x=14√3 and y=28

Step-by-step explanation:

This is a 30, 60, 90 triangle. We have to multiply 14 by 2 to get the hypotenuse. In this case, the hypotenuse is y. 14x2 is equal to 28. Therefore, y=18. The long side is equal to the small side times the √3. Therefore, x is equal to 14√3. Therefore, x=14√3 and y=28.

If this has helped you please mark as brainliest

4 0

3 years ago

A text message plan costs $9 per month plus $0.39 per text. Find the monthly cost for x text message The monthly cost of x messa

Sergio039 [100]

Answer:

The monthly cost = $9 + $0.39x

Step-by-step explanation:

4 0

2 years ago

How to find the derivative of cos^2x? i seem to be confused.

slamgirl [31]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2927231

————————

You can actually use either the product rule or the chain rule for this one. Observe:

• Method I:

y = cos² x

y = cos x · cos x

Differentiate it by applying the product rule:

$\mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}(cos\,x\cdot cos\,x)}\\\\\\
\mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}(cos\,x)\cdot cos\,x+cos\,x\cdot \dfrac{d}{dx}(cos\,x)}$

The derivative of cos x is – sin x. So you have

$\mathsf{\dfrac{dy}{dx}=(-sin\,x)\cdot cos\,x+cos\,x\cdot (-sin\,x)}\\\\\\
\mathsf{\dfrac{dy}{dx}=-sin\,x\cdot cos\,x-cos\,x\cdot sin\,x}$

$\therefore~~\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=-2\,sin\,x\cdot cos\,x}\end{array}}\qquad\quad\checkmark$

—————

• Method II:

You can also treat y as a composite function:

$\left\{\!
\begin{array}{l}
\mathsf{y=u^2}\\\\
\mathsf{u=cos\,x}
\end{array}
\right.$

and then, differentiate y by applying the chain rule:

$\mathsf{\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}}\\\\\\
\mathsf{\dfrac{dy}{dx}=\dfrac{d}{du}(u^2)\cdot \dfrac{d}{dx}(cos\,x)}$

For that first derivative with respect to u, just use the power rule, then you have

$\mathsf{\dfrac{dy}{dx}=2u^{2-1}\cdot \dfrac{d}{dx}(cos\,x)}\\\\\\
\mathsf{\dfrac{dy}{dx}=2u\cdot (-sin\,x)\qquad\quad (but~~u=cos\,x)}\\\\\\
\mathsf{\dfrac{dy}{dx}=2\,cos\,x\cdot (-sin\,x)}$

and then you get the same answer:

$\therefore~~\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=-2\,sin\,x\cdot cos\,x}\end{array}}\qquad\quad\checkmark$

I hope this helps. =)

Tags: <em>derivative chain rule product rule composite function trigonometric trig squared cosine cos differential integral calculus</em>

3 0

3 years ago

What is the length of line segment FE?

Jobisdone [24]

Answer:

(7,-1)because it is problem of vector lesson so that we have to use formula of vector

6 0

3 years ago

Read 2 more answers

GIVING BRAIN PLZ DO ASAP

Alekssandra [29.7K]

Answer: