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3 years ago

Solve for y. A. 10 B.12 C. 15 D. 18

Mathematics

2 answers:

LiRa [457]3 years ago

8 0

<h2>Answer:</h2>

Option A is the correct answer.

The value of y is:

A. 10

<h2>Step-by-step explanation:</h2>

From the figure we could see that:

3x=90°

so on dividing both side of the equation by 3 we obtain:

x=30°

and we see that ∠DBC=90°

Hence, we get:

2x+3y=90

on putting the value of x in the expression we get:

2×30+3y=90

i.e. 60°+3y=90°

⇒ 3y=90°-60°

⇒ 3y=30°

on dividing both side of the equation by 3 we obtain:

y=10°

Hence, the value of y is:

A. 10

djverab [1.8K]3 years ago

4 0

Y=10°
If we're solving for the ninety degree angle, which I seem to doubt, then if 3x=90 then 2x=60 giving a missing total of 30. 30/3 equals 10

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3 years ago

Given the functionf ( x ) = x^2 + 7 x + 10/ x^2 + 9 x + 20

vladimir1956 [14]

<em>x = -4 is a vertical asymptote for the function.</em>

<h2>Explanation:</h2>

The graph of $y=f(x)$ is a vertical has an asymptote at $x=a$ if at least one of the following statements is true:

$1) \ \underset{x\rightarrow a^{-}}{lim}f(x)=\infty\\ \\ 2) \ \underset{x\rightarrow a^{-}}{lim}f(x)=-\infty \\ \\ 3) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty \\ \\ 4) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty$

The function is:

$f(x)=\frac{x^2+7x+10}{x^2+9x+20}$

First of all, let't factor out:

$f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20} \\ \\ f(x)=\frac{x(x+5)+2(x+5)}{x(x+5)+4(x+5)} \\ \\ f(x)=\frac{(x+5)(x+2)}{(x+5)(x+4)} \\ \\ f(x)=\frac{(x+2)}{(x+4)}, \ x\neq 5$

From here:

$\bullet \ When \ x \ approaches \ -4 \ on \ the \ right: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(-4^{+}+2)}{(-4^{+}+4)} \\ \\ \\ The \ numerator \ is \ negative \ and \ the \ denominator \\ is \ a \ small \ positive \ number. \ So: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=-\infty$

$\bullet \ When \ x \ approaches \ -4 \ on \ the \ left: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(-4^{-}+2)}{(-4^{-}+4)} \\ \\ \\ The \ numerator \ is \ a \ negative \ and \ the \ denominator \\ is \ a \ small \ negative \ number \ too. \ So: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=+\infty$