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2 years ago

4X-y=6 3X+2y=21 Commomn core

Mathematics

1 answer:

Alex73 [517]2 years ago

3 0

Assuming you need to find the solution, it's x=3 and y=6
I used elimination

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Which graph shows the solution to the following equation?<br><br> 2x-5/x^2 = -4

Romashka [77]

Answer:

THE THIRD ONE

Step-by-step explanation:

4 0

3 years ago

Calculate the angle of depression from the

s344n2d4d5 [400]

Answer:

angle of depression ≈ 53.8°

Step-by-step explanation:

the angle of depression is the measure of the angle from the horizontal downwards from the top of the flag pole.

this angle is alternate to ∠ A and is congruent to ∠ A

using the sine ratio in the right triangle

sin A = $\frac{opposite}{hypotenuse}$ = $\frac{25}{31}$ , then

∠ A = $sin^{-1}$ ( $\frac{25}{31}$ ) ≈ 53.8° ( to 1 d.p )

Angle of depression ≈ 53.8°

6 0

1 year ago

Find the exact value of each trigonometric function for the given angle θ.

Kay [80]

Answer:

$\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=\dfrac{2}{\sqrt{3}}.$

Step-by-step explanation:

The given angle is 240 degrees.

We need to find the exact value of each trigonometric function for the given angle θ.

Since $\theta=240$ , it means θ lies in 3rd quadrant. In 3d quadrant only tan and cot are positive.

$\sin (240^\circ)=\sin (180^\circ+60^\circ)=-\sin (60^\circ)=-\dfrac{\sqrt{3}}{2}$

$\cos (240^\circ)=\cos (180^\circ+60^\circ)=-\cos (60^\circ)=-\dfrac{1}{2}$

$\tan (240^\circ)=\tan (180^\circ+60^\circ)=\tan (60^\circ)=\sqrt{3}$

$\cot (240^\circ)=\cot (180^\circ+60^\circ)=\cot (60^\circ)=\dfrac{1}{\sqrt{3}}$

$\sec (240^\circ)=\sec (180^\circ+60^\circ)=-\sec (60^\circ)=-2$

$\csc (240^\circ)=\csc (180^\circ+60^\circ)=-\csc (60^\circ)=-\dfrac{2}{\sqrt{3}}$

Therefore, $\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=-\dfrac{2}{\sqrt{3}}.$