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Delicious77 [7]
3 years ago
9

Which is the solution set of the inequality -15y+9<-36

Mathematics
1 answer:
liraira [26]3 years ago
6 0

Answer:

-15y+93\:\\ \:\mathrm{Interval\:Notation:}&\:\left(3,\:\infty \:\right)\end{bmatrix}

Therefore, the first choice is correct.

The graph of the inequality is also attached below.

Step-by-step explanation:

Considering the inequality

-15y+9

solving

-15y+9

\mathrm{Subtract\:}9\mathrm{\:from\:both\:sides}

-15y+9-9

-15y

\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}

\left(-15y\right)\left(-1\right)>\left(-45\right)\left(-1\right)

15y>45

\mathrm{Divide\:both\:sides\:by\:}15

\frac{15y}{15}>\frac{45}{15}

y>3

In other words,

-15y+93\:\\ \:\mathrm{Interval\:Notation:}&\:\left(3,\:\infty \:\right)\end{bmatrix}

Therefore, the first choice is correct.

The graph of the inequality is also attached below.

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If the least value of n is 5, which inequality best shows all the possible values of n? n ≥ 5 n ≤ 5 n &lt; 5 n &gt; 5
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3 years ago
If sinx = p and cosx = 4, work out the following forms :<br><br><br>​
Kay [80]

Answer:

$\frac{p^2 - 16} {4p^2 + 16} $

Step-by-step explanation:

I will work with radians.

$\frac {\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)} {[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]}$

First, I will deal with the numerator

$\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)$

Consider the following trigonometric identities:

$\boxed{\cos\left(\frac{\pi}{2}-x \right)=\sin(x)}$

$\boxed{\sin\left(\frac{\pi}{2}-x \right)=\cos(x)}$

\boxed{\sin(-x)=-\sin(x)}

\boxed{\cos(-x)=\cos(x)}

Therefore, the numerator will be

$\sin^2(x)-\sin(x)-\cos^2(x)+\sin(x) \implies \sin^2(x)- \cos^2(x)$

Once

\sin(x)=p

\cos(x)=4

$\sin^2(x)-\cos^2(x) \implies p^2-4^2 \implies \boxed{p^2-16}$

Now let's deal with the numerator

[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]

Using the sum and difference identities:

\boxed{\sin(a \pm b)=\sin(a) \cos(b) \pm \cos(a)\sin(b)}

\boxed{\cos(a \pm b)=\cos(a) \cos(b) \mp \sin(a)\sin(b)}

\sin(\pi -x) = \sin(x)

\sin(2\pi +x)=\sin(x)

\cos(2\pi-x)=\cos(x)

Therefore,

[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)] \implies [\sin(x)+\cos(x)] \cdot [\sin(x)\cos(x)]

\implies [p+4] \cdot [p \cdot 4]=4p^2+16p

The final expression will be

$\frac{p^2 - 16} {4p^2 + 16} $

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