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Dvinal [7]
2 years ago
8

Using the table, what is the value for the RANGE when the DOMAIN is 2?

Mathematics
1 answer:
user100 [1]2 years ago
5 0

Answer:

A) -1

Step-by-step explanation:

Domain is x value and range is y value

So when x = 2, then y = -1

Answer:

A) -1

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What are you asking? If it’s true or false it’s TRUE?
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2 years ago
A rectangular board has an area of 648 square centimeters. The triangular part of the board has an area of 162 square centimeter
maksim [4K]

Answer:

25%.

Step-by-step explanation:

Let E be the event that the dart lands inside the triangle.

We have been given that a rectangular board has an area of 648 square centimeters. The triangular part of the board has an area of 162 square centimeters.

We know that probability of an event represents the chance that an event will happen.

\text{Probability}=\frac{\text{Favorable no. of events}}{\text{Total number of possible outcomes}}

\text{Probability that dart lands inside the triangle}=\frac{\text{Area of triangle}}{\text{Area of rectangle}}

\text{Probability that dart lands inside the triangle}=\frac{162}{648}

\text{Probability that dart lands inside the triangle}=0.25

Convert into percentage:

0.25\times 100\%=25\%

Therefore, the probability that dart lands inside the triangle is 25%.

7 0
3 years ago
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kondaur [170]
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7 0
2 years ago
The absolute value function, f(x) = |x + 2|, is shown.
Anit [1.1K]

Answer:

Incomplete question

Step-by-step explanation:

4 0
2 years ago
If you wish to warm 40 kg of water by 18 ∘C for your bath, find what the quantity of heat is needed. Express your answer in calo
Harman [31]

Answer:

Quantity of heat needed (Q) = 722.753 × 10³

Step-by-step explanation:

According to question,

Mass of water (m) = 40 kg

Change in temperature ( ΔT) = 18°c

specific heat capacity of water = 4200 j kg^-1 k^-1

The specific heat capacity is the amount of heat required to change the temperature of 1 kg of substance to 1 degree celcius or 1 kelvin .

So, Heat (Q) = m×s×ΔT

Or,          Q = 40 kg × 4200 × 18

or,           Q = 3024 × 10³ joule

Hence, Quantity of heat needed (Q) = 3024 × 10³ joule    

In calories 4.184 joule = 1 calorie

So, 3024 × 10³ joule = 722.753 × 10³

6 0
3 years ago
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