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3 years ago

SAS, SSS, ASA, OR there is not enough information to determine congruence. please answer this asap !

Mathematics

1 answer:

butalik [34]3 years ago

3 0

There’s not enough info — you only have SS.

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Find an expression for the area enclosed by quadrilateral ABCD below.

Helga [31]

∆ABD is right angled hence area:-

$\\ \sf\longmapsto \dfrac{1}{2}bh$

$\\ \sf\longmapsto \dfrac{1}{2}(4x)(3x)$

$\\ \sf\longmapsto \dfrac{1}{2}(12x^2)$

$\\ \sf\longmapsto 6x^2$

There is only one option containing 6x^2 i.e Option D.

Hence without calculating further

Option D is correct

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3 years ago

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Artyom0805 [142]

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3 years ago

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X -1 < -4 What is ittttttttt

Kamila [148]

Answer:

x<-3

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3 years ago

D(x)=(x^2-12x+20)/(3x)

krok68 [10]

Answers:

Vertical asymptote: x = 0

Horizontal asymptote: None

Slant asymptote: (1/3)x - 4

Explanation:

d(x) = $\frac{x^{2}-12x+20}{3x}$

= $\frac{(x-2)(x - 10)}{3x}$

Discontinuities: (terms that cancel out from numerator and denominator):

Nothing cancels so there are NO discontinuities.

Vertical asymptote (denominator cannot equal zero):

3x ≠ 0

÷3 ÷3

x ≠ 0

So asymptote is to be drawn at x = 0

Horizontal asymptote (evaluate degree of numerator and denominator):

degree of numerator (2) > degree of denominator (1)

so there is NO horizontal asymptote but slant (oblique) must be calculated.

Slant (Oblique) Asymptote (divide numerator by denominator):

(1/3)x - 4
3x) x² - 12x + 20
x²
-12x
-12x
20 (stop! because there is no "x")

So, slant asymptote is to be drawn at (1/3)x - 4

6 0

3 years ago