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vaieri [72.5K]
3 years ago
9

Help! Based on the given information, what can you conclude and why?

Mathematics
1 answer:
DochEvi [55]3 years ago
6 0

Answer: \triangle HIJ \cong \triangle LKJ

Step-by-step explanation:

Given: ∠H ≅∠L and line HJ congruent to Line JL.

Since, In triangles HIJ and LKJ,

∠H ≅ ∠L

HJ ≅ JL

And, ∠HJI ≅ ∠ LJK

Therefore, By ASA postulate,

Δ HIJ ≅ Δ LKJ

This is why Option first is correct.

Note: Option third ΔHIJ≅ΔJKL is not correct,

Because,  HI≠JK



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Galina-37 [17]

The answer to this question is C.

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3 years ago
to design the interior surface of a huge stainless-steel tank, you revolve the curve y=x^2 on the interval 0x4 about the y-axis.
Oksana_A [137]

Answer:

W = 21.44*10⁶J

Step-by-step explanation:

Given

y = x²    (0 < x < 4)

γ = 10000 N/m³

W = ?

If we apply

y = 4² = 16 = h

y =  x²  ⇒  x = √y

then

V = π*(√y)²*dy = π*y*dy

W = (γ*V) *(16-y) = γ*π*y*dy*(16-y)

⇒  W = γ*π*∫y*(16-y)dy = γ*π*(8y²-(y³/3))

finally we obtain  (0 < y < 16)

W = γ*π*(8y²-(y³/3)) = 10000*π*(8*16²-(16³/3)) = 21.44*10⁶J

7 0
3 years ago
13<img src="https://tex.z-dn.net/?f=13%5Cfrac%7B1%7D%7B4%7D%20%2B2n%3D27%5Cfrac%7B3%7D%7B4%7D" id="TexFormula1" title="13\frac{1
Len [333]

Answer:

7 1/8

Step-by-step explanation:

Assuming that the first 13 was a mistake then the equation would be 13 1/2 + 2n = 27 3/4

subtrace 13 1/2 from 27 3/4

14 1/4

divide by 2

7 1/8

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3 years ago
Consider the relationship between the words pine and tree Which answer choice contains a pair of words with the same relationshi
BaLLatris [955]

Answer:

Maybe it's C.

Step-by-step explanation:

4 0
4 years ago
Trigonometric question, 30 points, will give brainliest.
zheka24 [161]

hmmm first off let's convert the √3 +i to trigonometric form, and then use De Moivre's root theorem, bearing in mind that √3 and i or 1i are both positive, meaning we're on the I Quadrant.

\bf (\stackrel{a}{\sqrt{3}}~,~\stackrel{b}{1i})\qquad \begin{cases} r=&\sqrt{(\sqrt{3})^2+1^2}\\ &\sqrt{3+1}\\ &2\\ \theta =&tan^{-1}\left( \frac{1}{\sqrt{3}}\right)\\\\ &tan^{-1}\left( \frac{\sqrt{3}}{3} \right)\\ &\frac{\pi }{6} \end{cases}~\hfill \implies ~\hfill 2\left[ cos\left( \frac{\pi }{6}\right) +i~sin\left( \frac{\pi }{6}\right) \right]

\bf ~\dotfill\\\\ \qquad \textit{power of two complex numbers} \\\\\ [\quad r[cos(\theta)+isin(\theta)]\quad ]^n\implies r^n[cos(n\cdot \theta)+isin(n\cdot \theta)] \\\\[-0.35em] ~\dotfill

\bf \left[ 2\left[ cos\left( \frac{\pi }{6}\right) +i~sin\left( \frac{\pi }{6}\right) \right] \right]^3\implies 2^3\left[ cos\left( 3\cdot \frac{\pi }{6}\right) +i~sin\left( 3\cdot \frac{\pi }{6}\right) \right] \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 8\left[cos\left( \frac{\pi }{2} \right) +i~sin\left( \frac{\pi }{2} \right) \right]~\hfill

3 0
4 years ago
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