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3 years ago

Help please help help please please please

Mathematics

1 answer:

AnnZ [28]3 years ago

3 0

Answer:

C. $\frac{5}{4}$

D. $\frac{6}{7}$

Step-by-step explanation:

C. $2 - \frac{12}{16}$

$= 2 - \frac{3}{4}$

$= \frac{8 - 3}{4}$

$= \frac{5}{4}$

D. $\frac{20}{22} \times \frac{33}{35}$

$= \frac{10}{11} \times \frac{33}{35}$

$= 10 \times \frac{3}{35}$

$= 2 \times \frac{3}{7}$

$= \frac{6}{7}$

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The lines are parallel

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3 years ago

S=2(lw+lh+wh) solve for w

wolverine [178]

we have

$s=2(lw+lh+wh)$

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3 years ago

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3 years ago

The course for a boat race starts at point A and proceeds in the direction of S52'E for 1 hour at 8 knots to point B and then in

KATRIN_1 [288]

Answer:

20.823 nautical miles

S20°E

Step-by-step explanation:

I assume you mean ° (degrees), not ' (minutes). There are 60 minutes in 1 degree.

S52°E means "south, 52° east", or 52° east of south.

S50°W means "south, 50° west", or 50° west of south.

1 knots = 1 nautical mile / hour, so the boat first travels 8 nautical miles from A to B, then 4.5 nautical miles from B to C, then finally back to A.

If we say A is at the origin, then the coordinates of B are:

(8 sin 52°, -8 cos 52°)

And the coordinates of C are:

(8 sin 52° − 4.5 sin 50°, -8 cos 52° − 4.5 cos 50°)

(2.857, -7.818)

So the distance from A to C is:

x = √(2.857² + (-7.818)²)

x ≈ 8.323

And the total distance of the race is:

d = 8 + 4.5 + 8.323

d = 20.823

The compass bearing from A to C is:

θ = atan(2.857 / 7.818)

θ ≈ S20°E

8 0

4 years ago