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Oliga [24]
2 years ago
13

Select three ratios that are equivalent to 7: 6.

Mathematics
1 answer:
scZoUnD [109]2 years ago
5 0

Answer:

ratio number 1-  14:12

ratio number 2- 21:18

ratio number 3- 28:24

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Evaluate the expression 5a + 5b. Let a = -6 and b = -5.
GrogVix [38]
<span>Evaluate the expression 5a + 5b. Let a = -6 and b = -5.

n = 5a + 5b
n = 5(-6) + 5(-5)
n = -30 + -25
n = -55

Hope this helps

</span>
8 0
3 years ago
Read 2 more answers
Find each measure m&lt;1, m&lt;2,m&lt;3
Leno4ka [110]

Answer:

option 2.

m1 = 75    ,   m2 = 129   ,   m3 = 100

Step-by-step explanation:

with the rule that the internal angles of a triangle add up to 180 ° we can calculate the missing angles

x + 46 + 29 = 180

x = 180 - 46 -29

x = 105

a flat angle has 180 °

m1 + 105 = 180

m1 = 180 - 105

m1 = 75

46 + 54 + y = 180

y = 180 - 46 -54

y = 80

80 = z + 29

z = 80 - 29

z = 51

as they are two crossed lines the angle is reflected from the opposite side

with that principle and knowing that the angle of a turn is 360 °, if we subtract the 2 known angles and divide it by 2 we will obtain the missing angle (m2)

m2 * 2 = 360 - 51 * 2

m2 = 258/2

m2 = 129

m2 = 29 + m3

129 = 29 + m3

m3 = 129 - 29

m3 = 100

4 0
3 years ago
A ship leaves port at 10 miles per hour, with a heading of N 35° W. There is a warning buoy located 5 miles directly north of th
Leno4ka [110]

The value of the angle subtended by the distance of the buoy from the

port is given by sine and cosine rule.

  • The bearing of the buoy from the is approximately <u>307.35°</u>

Reasons:

Location from which the ship sails = Port

The speed of the ship = 10 mph

Direction of the ship = N35°W

Location of the warning buoy = 5 miles north of the port

Required: The bearing of the warning buoy from the ship after 7.5 hours.

Solution:

The distance travelled by the ship = 7.5 hours × 10 mph = 75 miles

By cosine rule, we have;

a² = b² + c² - 2·b·c·cos(A)

Where;

a = The distance between the ship and the buoy

b = The distance between the ship and the port = 75 miles

c = The distance between the buoy and the port = 5 miles

Angle ∠A = The angle between the ship and the buoy = The bearing of the ship = 35°

Which gives;

a = √(75² + 5² - 2 × 75 × 5 × cos(35°))

By sine rule, we have;

\displaystyle \frac{a}{sin(A)} = \mathbf{ \frac{b}{sin(B)}}

Therefore;

\displaystyle sin(B)= \frac{b \cdot sin(A)}{a}

Which gives;

\displaystyle sin(B) = \mathbf{\frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }}

\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx 37.32^{\circ}

Similarly, we can get;

\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx \mathbf{ 142.68^{\circ}}

The angle subtended by the distance of the buoy from the port, <em>C</em> is therefore;

C ≈ 180° - 142.68° - 35° ≈ 2.32°

By alternate interior angles, we have;

The bearing of the warning buoy as seen from the ship is therefore;

Bearing of buoy ≈ 270° + 35° + 2.32° ≈ <u>307.35°</u>

Learn more about bearing in mathematics here:

brainly.com/question/23427938

5 0
2 years ago
?? can someone help plzzz
Stolb23 [73]

Answer:52,425


Step-by-step explanation: If you add all of the positive assets you end up with 15,050 but if you add 71,975 you get exactly 87,025 but when you add the liabilities you get -19,550 when you add those two numbers you get 52,425


Hope this helps!


3 0
3 years ago
I need help with 1,2,3 please
Len [333]

Answer:

1. Negative

2. Undefined

3. Zero

Step-by-step explanation:

6 0
2 years ago
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