Can anyone help me out?

adelina 88 [10]

ΔABD and ΔACD are similar (By A-A-A) and The length of DA is 6.

The triangles where all the corresponding sides of the two triangles are in equal proportion are called similar triangles.

In triangle ΔABC ∠BAC=90°

Let ∠ABC=x

then ∠ACB=90°-x

In triangle ΔADC, ∠ADC =90°

∠ACB=∠ACD=90°-x

∠DAC= 180°-(∠ADC+∠ACD)= 180°-(90°+90°-x)= x

In triangle ΔADB , ∠ADB =90°

∠ABD=x

∠BAD=90°-x

Between triangles ΔABD and ΔACD

∠ABD=∠DAC (=X)

∠BAD∠ACD (=90°-x)

∠ADB=∠ADC (=90°)

from the above, it is clear that triangles ΔABD nad ΔACD is similar. (By A-A-A)

So sides are in equal proportion in 2 triangles,

AD/DC= BD/AD

⇒AD²= BD*DC

⇒AD²=9*4

⇒AD²=36

⇒AD=6

⇒DA=6

Therefore triangles ΔABD and ΔACD are similar (By A-A-A) and The length of DA is 6.

Learn more about similar triangle

here: brainly.com/question/14285697

#SPJ10

5 0

3 years ago

Is the integral ſ A. dr path independent if A = (2x - y)i + (x + y)j?

evablogger [386]

The integral is path-independent if we can find a scalar function f such that grad(f ) = A. This requires

$\dfrac{\partial f}{\partial x}=2x-y$

$\dfrac{\partial f}{\partial y}=x+y$

Take the first PDE and integrate both sides with respect to x to get

$f(x,y)=x^2-xy+g(y)$

where g is assumed to be a function of y alone. Differentiating this with respect to x gives

$\dfrac{\partial f}{\partial y}=-x+\dfrac{\mathrm dg}{\mathrm dy}=x+y\implies\dfrac{\mathrm dg}{\mathrm dy}=2x+y$

which would mean g is *not* a function of only y, but also x, contradicting our assumption. So the integral is path-dependent.

Parameterize the unit circle (call it C) by the vector function,

$\mathbf r(t)=\cos t\,\mathbf i+\sin t\,\mathbf j$

with t between 0 and 2π.

Note that this parameterization takes C to have counter-clockwise orientation; if we compute the line integral of A over C, we can multiply the result by -1 to get the value of the integral in the opposite, clockwise direction.

Then

$\mathrm d\mathbf r=-\sin t\,\mathbf i+\cos t\,\mathbf j$

and the (counter-clockwise) integral over C is

$\displaystyle\int_C\mathbf A\cdot\mathrm d\mathbf r$

$\displaystyle=\int_0^{2\pi}((2\cos t-\sin t)\,\mathbf i+(\cos t+\sin t)\,\mathbf j)\cdot(-\sin t\,\mathbf i+\cos t\,\mathbf j)\,\mathrm dt$

$\displaystyle=\int_0^{2\pi}1-\sin t\cos t\,\mathrm dt=2\pi$

and so the integral in the direction we want is -2π.

By the way, that the integral doesn't have a value of 0 is more evidence of the fact that the integral is path-dependent.

5 0

3 years ago

EXPRESS in power notation

lora16 [44]

Answer:

Solution given:

1:

-1/8= $\frac{-1³}{2³}=(\frac{-1³}{2³}=-1³*2^{-3}=-2^{-3}$

2. -64/27

= $\frac{-4^{3}}{3³}=\frac{-4³}{3³}=(\frac{-4}{3})^{3}$

Express in rational number

1. (-3/2)=-3/2*2/2=-(3*2)*(2/2)=-6/4

2. (1/5)=1/5*5/5=5/25

and

(-4/3)³(2/5)-⁴ ÷ (7/4)

(-4³/3³)(2-⁴/5-⁴)÷7/4

(-64/27)(5⁴/2⁴)÷7/4

(-64/27)(625/16)÷7/4

(-64*625/(27*16))*4/7

-2500/27*4/7

-10000/169

(-100/13)²

7 0

3 years ago

What is the value of the expression –7 – 5 + 4?

balandron [24]

This is A (-16)... When looking for this use the fomula PEMDAS ( Parenthesis, Exponents, Multiply, Add, and Subtract)....... 5+4=9 and -7-9=-16... so the answer is A

6 0

3 years ago

Read 2 more answers

What is the answer to the problem

vovikov84 [41]

Answer: 6

Step-by-step explanation: DE is 4+2 so EF would be the same if it is equal.

3 0

3 years ago