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3 years ago

Help leave in simplest radical form

Mathematics

1 answer:

nikklg [1K]3 years ago

8 0

Hi. Hope it has helped you.

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Solve for the missing side. Round to the nearest tenth. х 51° 12

alisha [4.7K]

9514 1404 393

Answer:

7.6

Step-by-step explanation:

The relevant relation is ...

Cos = Adjacent/Hypotenuse

cos(51°) = x/12

x = 12·cos(51°) ≈ 7.552

x ≈ 7.6

6 0

3 years ago

For a certain map, the scale shows that 1 cm on the map equals 2.50 km on earth's surface. two towns on the map are 4.75 cm apar

Julli [10]

In this problem, to find the answer we have to setup a series of ratios that relate the scale to real life distance. We know that 1cm = 2.50km, so that ratio would be 1cm/2.5km. For two towns that are 4.75cm apart on the map, we set a ration of 4.75cm/x km, where x is the actual distance. Now we set the ratios equal to each other and solve for x. 1/2.5=4.75/x where x = 4.75*2.5/1 = 11.875 and rounding up we get 11.88 km. The two towns are actually 11.88 km apart from each other.

3 0

3 years ago

Please help!

mr Goodwill [35]

Answer:

1. $\sqrt{74}$ ft 2. $50\sqrt{2}$ yards 3. $5\sqrt{35}$ units.

Step-by-step explanation:

Pythagorean's Theorem $a^2 + b^2 = c^2$

A and b are both side lengths, c is the hypotenuse.

7^2 + 5^2 = 74

sqrt(74) is the answer for 1.

30^2 - 5^2 = 875

Simplify the radical. sqrt(875) --> 5sqrt(35). The answer for three.

There's a rule in geometry that says a diagonal of a square is the same length as taking a side length times the square root of two. There's your answer for two.

7 0

3 years ago

Find the radius and the diameter to the nearest hundredth when c = 6π yd. use 3.14 for π

Natali [406]

Radius is 3 And diameter 3×2

8 0

4 years ago

Which ordered pair makes both inequalities true?

kvasek [131]

Answer:

(3, 0)

Step-by-step explanation:

Given the inequality y > -2x + 3 and y ≤ x - 2

The graph of the inequalities are plotted using the geogebra graphing online calculator.

The portion of the graph that is shaded with dark blue, represents the portion that supports the equation.

All the ordered pair points given in the question are also labelled in the graph.

From the graph we can see that only point (3, 0) falls in the area that supports the equation. Hence (3, 0) makes both inequalities true.

6 0

3 years ago