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Rufina [12.5K]
3 years ago
5

Solve the following expression when a = 4 and c = 3 4a + 12 – 2c + c Order of operation

Mathematics
2 answers:
lbvjy [14]3 years ago
8 0

Answer:

19

Step-by-step explanation:

4×4= 16+12= 28

2×3= 6+3= 9

28-9= 19

lianna [129]3 years ago
8 0
4x4 = 16 16 + 12 = 28 2 x 3 = 6 6 + 3 = 9

28 - 9 = 19
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One more question on this test guys. please answer as fast as possible. If you do,
GaryK [48]

1. The given rectangular equation is x=2.

We substitute x=r\cos \theta.

r\cos \theta=2

Divide through by \cos \theta

r=\frac{2}{\cos \theta}

r=2}\sec \theta

\boxed{x=2\to r=2\sec \theta}

2. The given rectangular equation is:

x^2+y^2=36

This is the same as:

x^2+y^2=6^2

We use the relation r^2=x^2+y^2

This implies that:

r^2=6^2

\therefore r=6

\boxed{x^2+y^2=36\to r=6}

3. The given rectangular equation is:

x^2+y^2=2y

This is the same as:

We use the relation r^2=x^2+y^2 and y=r\sin \theta

This implies that:

r^2=2r\sin \theta

Divide through by r

r=2\sin \theta

\boxed{x^2+y^2=2y\to r=2\sin \theta}

4. We have x=\sqrt{3}y

We substitute y=r\sin \theta and x=r\cos \theta

r\cos \theta=r\sin \theta\sqrt{3}

This implies that;

\tan \theta=\frac{\sqrt{3}}{3}

\theta=\frac{\pi}{6}

\boxed{x=\sqrt{3}y\to \theta=\frac{\pi}{6}}

5. We have x=y

We substitute y=r\sin \theta and x=r\cos \theta

r\cos \theta=r\sin \theta

This implies that;

\tan \theta=1

\theta=\frac{\pi}{4}

\boxed{x=y\to \theta=\frac{\pi}{4}}

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