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3 years ago

PLEAASSSSSSSEEEE HELLPPPPPP !!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

tatyana61 [14]3 years ago

5 0

Answer:

0 to 15

14 to 8

12 to 9

2 to 14

Step-by-step explanation:

Please give me brainliest, I need it! :D

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Can someone help me draw this out to visualize it?

Vedmedyk [2.9K]

Answer:

Solution given:

South distance :base[b]=80milea

East distance :perpendicular [p]=35miles

Now

<S=?

we have

$\tan( \alpha ) = \frac{p}{b} = \frac{35 }{80}$

$\alpha = \tan {}^{ - 1} ( \frac{7}{16} )$

$\alpha = 23.629°$ =24°

24° bearing should be taken from south airport to East airport.

3 0

3 years ago

Hhhheeeeellllllpppp no 13 and 14

laila [671]

Answer:

13. B. 24 cm²

14. y = 5⅝ (none of the options is correct)

Step-by-step explanation:

13. The figure is a trapezoid

Area of the trapezoid = ½(a + b)h

Where,

a = 3 cm

b = 5 cm

h = 6 cm

Area = ½(3 + 5)6

= ½(8)6

= 4*6

= 24 cm²

14. Area of the triangle = ½*b*h

The area is given to be 45 cm²

base (b) = 16 cm

height (h) = y

Plug in the values and solve for y

45 = ½*16*y

45 = 8y

Divide both sides by 8

45/8 = 8y/8

45/8 = y

5⅝ = y

y = 5⅝ (none of the options is correct)

5 0

3 years ago

Using the measurements given below, find the hypotenuse of the triangle.

zloy xaker [14]

Answer:

In the triangle shown, AB = 11, BC = 61. Find AC. Right Triangle ABC v2. 3,600; 3,842; 60; 62. 4. Using the following measurements, find the length of the leg of the right triangle. leg = 5

Step-by-step explanation:

Not sure if that's helpful, but hope it is.

5 0

3 years ago

One day at 3:00 a.m., the temperature was -13°F in Kodiak, Alaska. At 10:00 a.m., the temperature was 22°F. What was the average

suter [353]

-13 to 0 = 13 degrees changed. 0 to 22 is 22 degrees changed. 13 + 22 = 35 degree change.

10-3 = 7.

So there was a change of 35 degrees across 7 hours

35/7 = 5 degrees per hour.

6 0

3 years ago

The velocity v and maximum height h of the water being pumped into the air are related by the equation v=\sqrt(2gh) where g is t

Whitepunk [10]

The velocity v and maximum height h of the water being pumped into the air are related by the equation

v= $\sqrt{2gh}$