Solve equation y+5=-12

Mathematics

2 answers:

densk [106]4 years ago

6 0

The answer to your problem it -17 Hope this helped

DochEvi [55]4 years ago

4 0

Hope this helps :).

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Solve for u. -5/6u = 15

Umnica [9.8K]

Answer:

u = -18

Step-by-step explanation:

$\frac{-5u}{6} =15$

Multiply both sides by 6.

-5u = 15 × 6

-5u = 90

Divide both sides by -5.

u = -18

3 0

2 years ago

How do I verify the identity: sec(t)+1/tan(t)=tan(t)/sec(t)-1

kompoz [17]

$\bf \textit{Pythagorean Identities}
\\ \quad \\
sin^2(\theta)+cos^2(\theta)=1
\\ \quad \\
1+cot^2(\theta)=csc^2(\theta)
\\ \quad \\
\boxed{1+tan^2(\theta)=sec^2(\theta)}\implies tan^2(\theta)=sec^2(\theta)-1\\\\
-----------------------------\\\\
\cfrac{sec(x)+1}{tan(x)}=\cfrac{tan(x)}{sec(x)-1}\\\\
-----------------------------\\\\$

$\bf \cfrac{sec(x)+1}{tan(x)}\cdot \cfrac{sec(x)-1}{sec(x)-1}\impliedby \textit{using the conjugate}\\\\
-----------------------------\\\\
recall\qquad \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-----------------------------\\\\
thus\qquad \cfrac{sec^2(x)-1^2}{tan(x)sec(x)-1}\implies \cfrac{sec^2(x)-1}{tan(x)sec(x)-1}
\\\\\\
\cfrac{\underline{tan^2(x)}}{\underline{tan(x)} sec(x)-1}\implies \cfrac{tan(x)}{sec(x)-1}$

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4 years ago

What is the value of y? 3√3 6√3 9√3 12√3

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