The answer is “It is one-half the area of a rectangle of length 4 units and width 3 units.” This is because the height of this triangle is 3 units and the base is 4 units so it will have the same area as half a rectangle with the same units.
Answer:
Length: 27 width:9
Step-by-step explanation:
27 + 27 + 9 + 9 = 72 and 27 is 3 x 9
Answer:
essentially, in these problems, youre finding the square root of each number. the square root should be a decimal thats in between two other numbers, and those will be the integers that the value of the square root is between. unless the value is exactly a number and a half, it will be closer to one integer than the other. the work is below.
square root of:
59= 7.68
68= 8.246
78= 8.83
93= 9.64
104= 10.198
124= 11.14
33= 5.74
185= 13.6
215= 14.66
these are just the square roots. but, as you can see, each number is between two other whole numbers on a number line. So, just write down what those numbers are.
7.68 is between 7 and 8
8.246 is between 8 and 9
8.83 is between 8 and 9
9.64 is between 9 and 10
10.198 is between 10 and 11
11.14 is between 11 and 12
5.74 is between 5 and 6
13.6 is between 13 and 14
14.66 is between 14 and 15
so, we know thw numbers that each decimal is between, but we have to find which one each is cloest to. in order to do that, just know that if the decimal is more than .5 its closer to the larger number, and its its less than .5, its closer to thw smaller number.
7.68 is closer to 8
8.246 is closer to 8
8.83 is closer to 9
9.64 is closer to 10
10.198 is closer to 10
11.14 is closer to 11
5.74 is closer to 6
13.6 is closer to 14
14.66 is closer to 15
Answer:
Step-by-step explanation:
The given equations are:
(1)
and 
(2)
Now, simply adding equation (1) and equation (2), we get
Now, substituting the value of x=6 in equation (2). we get
Thus, the values of x and y are 6 and 1 respectively.
In order to solve the system of solution, beau can see that in both the given equations, coefficient of y is same and is with opposite sign, thus simply adding both the equations will eliminate y and she will get the value of x and then substituting the value of x in any of the equations, she can get the value of y. With this, she can get closer to the solution.
