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bija089 [108]
2 years ago
11

A number decreased by 4 is greater than 13 solve pls

Mathematics
1 answer:
eimsori [14]2 years ago
4 0

Answer:  

Year 10 Interactive Maths - Second Edition

Problem Solving

Linear equations are often used to solve practical problems that have an unknown quantity. We use a suitable pronumeral to represent the unknown quantity, translate the information given in the problem into an equation, and then solve the equation using the skills acquired earlier in this chapter.

Example 11

If a number is increased by 8, the result is 25. Find the number.

Solution:

Let x be the number.  Increasing x by 8 gives x + 8, which we are told is 25.  Therefore, x + 8 = 25.  Subtract 8 from both sides to find x = 17.  So, the number is 17.

Step-by-step explanation:

I hope this helps!!!

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Yuki888 [10]

Answer:

try 36?

Step-by-step explanation:

5 0
3 years ago
April had
Bumek [7]
First thing First. You must find have a common denominator. If you want to find it then you do this

3: 3 6 9 12 15 18 21
7: 7 14 21 28

Once you found a number they have in common (21) You do this next

3/7 * 3/3 = 9/21
2/3 * 7/7 = 14/21

Now the next thing you do is subtract your numerators but not the denominators

14-9=5

5/12

you cant simplified so your done! Hope this helps!!!!  <span />
3 0
3 years ago
How do you cauculate net change ​
Travka [436]

Answer:

You can calculate net change by subtracting the current day's closing price for an asset from the closing price of the previous day!

Step-by-step explanation:

(What net change is) The net change theorem states that when a quantity changes, the final value equals the initial value plus the integral of the rate of change. Net change can be a positive number, a negative number, or zero.

NOT MY ANSWER OR WHATEVER I GOT IT FROM G0OGLE

7 0
3 years ago
Write the following percentages as a fraction in simplest form
alisha [4.7K]
Answer: 241/4%
hope this helped
4 0
3 years ago
Someone help with this
elena-14-01-66 [18.8K]

The quadrants are I, II, III, and IV, (meaning 1, 2, 3, and 4). The first quadrant is in the upper right, which has both positive x and positive y values. The second quadrant is the upper left, which has negative x and positive y values. The third quadrant is the lower left, which has both negative x and y values. The fourth quadrant is the lower right, which has positive x and negative y values. Using this knowledge and our positive and negative signs, the following are the answers to questions 1-6.

1) (-4, -2) - Quadrant III, both x and y are negative

2) (0, -7) - This point is actually on the y-axis. The x value is 0 and the y value is -7, so the graph is 7 units down from the origin on the y-axis.

3) (0,0) - This point is the origin, or where the x and y axes cross (the middle of the graph).

4) (6, -9) - This point is in Quadrant IV, because it has a positive x value and a negative y value

5) (3,5) - This point is in Quadrant I, because both the x and y values are positive.

6) (8,0) - This point is on the x-axis. The y-value is zero, so this point is 8 units to the right of the origin between Quadrant I and Quadrant IV.

Using the knowledge presented above, to graph the points given to you in the second part of the problem, first you can figure out what quadrant or part of the graph the point is on. Then, you can count the number of units (squares on the graph) in the right direction (remember that up is positive on the y-axis and down is negative, and to the right is positive on the x-axis and to the left is negative) in order to plot the points. Then, you must connect the points that correspond to the same figure in order to create the figures.

Please comment if you have any questions!

Hope this helps!

7 0
3 years ago
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