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3 years ago

14

Graph r < – 4 orr > 3?

1 answer:

Nadya [2.5K]3 years ago

3 0

Answer:

Graph r < – 4 orr > 3?

Step-by-step explanation:

Graph r < – 4 orr > 3?

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Under what circumstances does the system of equations Qx+Ry=S and Y=Tx+S have infinitely many solutions?

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From these -Tx+y=S. If -T=Q/R, then y=-Qx/R+S, so Ry=-Qx+RS, Qx+Ry=RS=S.
If R is not equal to 1, or S is non-zero, the equations are inconsistent, so there would be no solutions.
If R=1 there are an infinite number of solutions given by Qx+y=S, or y=S-Qx or y=S+Tx.
If S=0, Qx+Ry=0 or y=-Qx/R or y=Tx.

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2 years ago

Cot^2x/cscx-1=1+sinx/sinx

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$\bf \textit{difference of squares}
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(a-b)(a+b) = a^2-b^2\qquad \qquad 
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$\bf \cfrac{\quad \frac{cos^2(x)}{sin^2(x)}\quad }{\frac{1}{sin(x)}-1}\implies \cfrac{\quad \frac{cos^2(x)}{sin^2(x)}\quad }{\frac{1-sin(x)}{sin(x)}}\implies \cfrac{cos^2(x)}{sin^2(x)}\cdot \cfrac{sin(x)}{1-sin(x)}
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2 years ago

I need help! I don't know how to get the answer or what it is. Can anyone help me it's for a test and I'm on my last attempt?

Greeley [361]

Answer:

See explanation

Step-by-step explanation:

(4)

Using the sine ratio in the right triangle

sinΘ = $\frac{opposite}{hypotenuse}$ = $\frac{7}{9}$ , thus

Θ = $sin^{-1}$ ( $\frac{7}{9}$ ) ≈ 51.1°

(5)

Using the tangent ratio in the right triangle

tanΘ = $\frac{opposite}{adjacent}$ = $\frac{4}{5}$ , thus

Θ = $tan^{-1}$ ( $\frac{4}{5}$ ) ≈ 38.7°

7 0

3 years ago