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Law Incorporation [45]
3 years ago
6

1.3.2 checkup - lessons learned

Mathematics
1 answer:
den301095 [7]3 years ago
5 0

Answer:

y=2/3x-4

Step-by-step explanation:

we can see that x goes up 1 for every 1.5y and if we start y at 0 x starts at -4 so if y is one than x has to be 2/3-4 becuase it is -4 + 2/3 for every 1 y goes up or one for every 1.5 y goes up. hope this answer was helpful.

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If f(x)=5x^3 and g(x)=x+1, find (f•g)(x)
alukav5142 [94]

Hello from MrBillDoesMath!

Answer:

5 x^3 + 15 x^2 + 15 x + 5 , none of the provided choices

Discussion:

f(x) = 5 x^3

g(x) = x+ 1  

=>

(f•g)(x) =

f(g(x)) =

f(x+1) =

5 * (x+1)^3 =

5 x^3 + 15 x^2 + 15 x + 5

which is none of the provided answers.

Thank you,

MrB

5 0
3 years ago
A movie club charges a one-time fee membership fee of $25. This allows members to purchase movies for $7 each. Lincoln is a memb
AleksAgata [21]

Answer:

he purchased 5 movies bro

Step-by-step explanation:

3 0
3 years ago
Plz help me with this
shtirl [24]
2x-7y=18 _(1)
-2x+4y=5

11y=23 , Y=23/11 sub in 1

2x-7×23/11=18

2x = 18×11 + 7×23

x = ( 198 + 161 )/2 = 359/2
5 0
4 years ago
56 times a number minus 13 is equal to 6 less than the number.
Eddi Din [679]

Answer:

<h2>Z = 19/56</h2>

<h2>OR Z = 0.34</h2>

Step-by-step explanation:

The steps in the photo above

I hope that is useful for you :)

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
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