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3 years ago

4 SPARSE BAYESIAN LEARNING

Mathematics

1 answer:

Scorpion4ik [409]3 years ago

7 0

Answer:

Answer:

WF=34

Step-by-step explanation:

Point I lies between points W and F.

It means point WI + IF = WF

SO WI + IF = WF

WF = 15x – 21

WI= 7x-3

IF= 2x+4

WI + IF = WF

7x-3 +2x+4= 15X-21

7x+2x-15x= -21-4+3

-6x = -22

X= -22/6

X= 11/3

So WF = 15x – 21

WI= 7x-3

IF= 2x+4

WI + IF = WF

7X-3 +2x+4= 15x-21

7x+2x-15x= -21-4+3

-6x = -22

X= -22/6

X= 11/3

So WF = 15x – 21

WF= 15(11/3) -21

WF= 55-21

WF=34

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In the figure below, m∠ABC = 57 degrees and m∠DAB =116 degrees. What is m∠ABC?

Allushta [10]

Answer:

∠ACB = 59°

Step-by-step explanation:

∠BAC = 180 - 116 = 64°

∠ACB = 180 - 64 - 57 = 59°

3 0

3 years ago

Evaluate the limit of

tekilochka [14]

Answer:

Option a is correct -- $\frac{9}{2}$

Step-by-step explanation:

If we put infinity directly into the expression we get ∞ + ∞ expression. In order to circumvent it we divide both numerator and denominator with the greatest exponent.

$\lim_{x \to \infty} \frac{9n^{3}+5n-2}{2n^{3}}$

Divide each term by n³

$=\lim_{x \to \infty} \frac{\frac{9n^{3}}{n^{3}}+\frac{5n}{n^{3}} -\frac{2}{n^{3}}}{\frac{2n^{3}}{n^{3}} }$

$=\lim_{x \to \infty}\frac{9(1)+\frac{5}{n^{2}}-\frac{2}{n^{3}}}{2(1)} } \\=\lim_{x \to \infty}\frac{9+\frac{5}{n^{2}}-\frac{2}{n^{3}}}{2}} \\\\by putting n = infinity \frac{5}{n^{2}} becomes 0 and\frac{2}{n^{3}} becomes 0 \\=\frac{9+0-0}{2} \\=\frac{9}{2}$