A veterinarian needs to know an animal’s weight in kilograms when treating them. If 20 pounds is about 9 kilograms and a dog weighs 30 pounds determine the dog’s weight in kilograms.

Answers

Answer 1

Answer: 1 kilo is approximately 2.2 pounds, so we can divide 30 by 2.2

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Find the measures of the supplementary angles that satisfy each case.The measure of the first angle is 45° more than the measure of the second.
(I will make you braniliest if it is correct)

Answers

Answer:

67.5°, 107.5°

Step-by-step explanation:

For supplementary angles, their sum equals 180°.

Let x be the first angle and y be the second angle, then

x + y = 180°.

It is given that x = y + 45°.

So x + y = 180°

substituting x into the equation, we have

y + 45° + y = 180°

simplifying, we have

2y + 45° = 180°

collecting like terms, we have

2y = 180° - 45°

2y = 135°

dividing through by 2, we have

y = 135°/2

y = 67.5°

Since y = 67.5°

then x = y + 45°

x = 67.5° + 45°

x = 107.5°

Final answer:

The measures of the supplementary angles that satisfy the given conditions are 67.5° and 112.5°.

Explanation:

Let's say the measure of the second angle is x. Since the measure of the first angle is 45° more than the measure of the second, we can express the first angle as the measure of x + 45°.

By the definition of supplementary angles, we know that the sum of the measures of two supplementary angles is 180°. Therefore, we can create the following equation: x + (x + 45) = 180.

Solving this equation gives us:

Add the expressions on the left side of the equation to get: 2x + 45 = 180.
Subtract 45 from both sides of the equation to isolate the term containing x: 2x = 135.
Divide both sides of the equation by 2 to solve for x: x = 67.5°.

So, "the second angle measures 67.5°, and the first angle, being x + 45°, measures 112.5°".

Learn more about Supplementary Angles here:

brainly.com/question/18164299

#SPJ12

I am not understanding this question.

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Answer:

boardgamegeek.com/boardgame/245655/kings-dilemma

Step-by-step explanation:

Please help! If it is right I will mark it brainliest!!!!A ball is thrown upward from the top of a building. The function below shows the height of the ball in relation to sea level, f(t), in feet, at different times, t, in seconds: f(t) = −16t2 + 48t + 160 The average rate of change of f(t) from t = 3 seconds to t = 5 seconds is feet per second.

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the answer is -80 :)

The answer is -80. I just finished taking this test and this question was right.

among a group of students 50 played cricket 50 played hockey and 40 played volleyball. 15 played both cricket and hockey 20 played both hockey and volleyball 15 played cricket and volley ball and 10 played all three. if every student played at least 1 game find the no of students and how many students played only cricket, only hockey and only volley ball

Answers

Answer:

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Explanation:

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

Answer:

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Explanation of the answer:

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

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An above-ground swimming pool in the shape of a cylinder has a diameter of 16 feet and a height of 5 feet. If the pool is filled with water to 1.5 inches from the top of the pool, what is the volume, to the nearest cubic foot, of the water in the pool?⇆