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Kryger [21]
3 years ago
6

Which constant could each equation be multiplied by to eliminate the x variable in this system of equations? 4x + 5y = 62 5x − 3

y = 22
Mathematics
1 answer:
qwelly [4]3 years ago
7 0

Answer:

Multiply equation 4x+5y=62 by 5 and 5x-3y=22 by 4

Step-by-step explanation:

Given:

Equations are 4x+5y=62,5x-3y=22

To find: constant by which each equation should be multiplied to eliminate the variable x in the given system of equations

Solution:

4x+5y=62...(i)\\5x-3y=22...(ii)

Multiply equation (i) by 5 and (ii) by 4 to get the following equations:

20x+25y=310\\20x-12y=88

On subtracting these equations, we get

(20x+25y)-(20x-12y)=310-88\\\\13y=222

So, the variable x gets eliminated

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Answer:

A≈345.58cm²

Step-by-step explanation:

Radius= 5cm and the Height= 6cm

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3 years ago
Jeremy ran 1500 meters in 4 minutes, 45 seconds. How fast did he run in meters per second?
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3 years ago
There are two numbers whose sum is 53. Three times the smaller number is equal to 19 more than the larger number. What are the n
julsineya [31]

18 and 35. The numbers whose sum 53 are 18 and 35.

The key to solve this problem is using a system of equations.

There are two numbers whose sum is 53. This number can be represented as x and y. So:

x + y = 53

Three times the smaller number is equal to 19 more than the larger. Let's set x as the smaller number and y the larger number. So:

3x = 19 + y

Clear y in both equations and let's use the equalization method to solve for x:

y = 53 - x and y = 3x - 19

Then,

53 - x = 3x - 19

53 + 19 = 3x + x ---------> 3x + x = 53 + 19 -------> 4x = 72

x = 72/4 = 18

To find y, let's substitute x = 18 in the equation x + y = 53

18 + y = 53 --------> y = 53 - 18

y = 35

6 0
3 years ago
if the sum of the first 60 positive integers is s, what is the sum of the first 120 integers in terms of s? a. 2s 3600 b. s^2 36
natima [27]
For the first 60 positive integers, a = 1, n = 60, l = 60.
Sn = n/2(a + l)
s = 60/2(1 + 60) = 30(61)

For the next 60 positive integer, a = 61, n = 60, l = 120
Sum = 60/2(61 + 120) = 30(61 + 120) = 30(61) + 30(120) = s + 3600

Sum of first 120 positive integers = s + s + 3600 = 2s + 3600
3 0
3 years ago
What would x any y be? <br> Simply your answer pls
Helen [10]
Look it up on google
5 0
3 years ago
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