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GrogVix [38]
3 years ago
9

Find the average rate of change

Mathematics
1 answer:
serious [3.7K]3 years ago
3 0

Answer:

-6

Step-by-step explanation:

kyo na ang bhala sa answer

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Find the measure of angle of hmg<br><br>PLS HELP AND SHOW WORK​
Anna11 [10]

Answer:

70 degrees

Step-by-step explanation:

Angle HMK = 50 degrees

Angles LMK, MLK and MKL are equal to 60 degrees, being in a perfectly congruent triangle which they all add up to 180 degrees.

Add 60 to 50 = 110

180 - 110 = 70 degrees, equal to angle HMG.

This is because the line that angles LMK, HMK and HMG are on is supplementary.

8 0
3 years ago
2. Celesta used to spend 90 minutes a day on the phone. Now she spends
algol13

Answer:

33%

Step-by-step explanation:

sorry my bad i did percentage increase.

7 0
4 years ago
Can you Please help me?
Alenkasestr [34]

Answer:

131 and 49

Step-by-step explanation:

Supplementary angle means 180° when all angles are added together.

First angle: (x + 82)

Second angle: x

(x + 82) + x = 180

2x + 82 = 180

2x = 98

x = 49

First angle: 49 + 82 = 131°

Second angle: 49°

8 0
4 years ago
Janet scored 200 and 400 points in the first 2 rounds of a computer game.
7nadin3 [17]

Answer:

B. (200+400)-250=p

Step-by-step explanation:

Let p be number of points Michael needs to score in 2nd game to catch up.

We have been given that Janet scored 200 and 400 points in the first 2 rounds of a computer game. So the total points scored by Janet in two games will be: (200+400) points.

We are also told that Michael had scored 250 points in the first round and he want to get the same total score as Janet.

So we can find the number of points Michael needs to score to catch up Janet by subtracting the number of points scored by Michael in 1st game from total number of points scored by Janet in two games. We can represent this information in an equation as:

(200+400)-250=p

Therefore, the equation (200+400)-250=p will help Michael to find the number of points he need to catch up Janet and option B is the correct choice.


3 0
4 years ago
Read 2 more answers
A homogeneous rectangular lamina has constant area density ρ. Find the moment of inertia of the lamina about one corner
frozen [14]

Answer:

I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)

Step-by-step explanation:

By applying the concept of calculus;

the moment of inertia of the lamina about one corner I_{corner} is:

I_{corner} = \int\limits \int\limits_R (x^2+y^2)  \rho d A \\ \\ I_{corner} = \int\limits^a_0\int\limits^b_0 \rho(x^2+y^2) dy dx

where :

(a and b are the length and the breath of the rectangle respectively )

I_{corner} =  \rho \int\limits^a_0 {x^2y}+ \frac{y^3}{3} |^ {^ b}_{_0} \, dx

I_{corner} =  \rho \int\limits^a_0 (bx^2 + \frac{b^3}{3})dx

I_{corner} =  \rho [\frac{bx^3}{3}+ \frac{b^3x}{3}]^ {^ a} _{_0}

I_{corner} =  \rho [\frac{a^3b}{3}+ \frac{ab^3}{3}]

I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)

Thus; the moment of inertia of the lamina about one corner is I_{corner} =\frac{\rho _{ab}}{3}(a^2+b^2)

7 0
3 years ago
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