Answer:
i think the first one but im not sure
im sorry if its wrong
Step-by-step explanation:
The size of the second application given the size of the first application and the expression ( x - 3.45 mb) for the size of the second application is 293.55 MB.
<h3>Equation</h3>
Let
- Size of the first application = x
- Size of the second application= x - 3.45 mb
For instance,
if the size of the first application is 297 MB
Size of the second application= x - 3.45 mb
= 297 MB - 3.45 MB
= 293.55 MB
Therefore, the size of the second application given the size of the first application and the expression for the size of the second application is 293.55 MB
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Number one is but number two is not
of more than 1 dot is on an “up and down” line then it’s not a function
Answer:
Step-by-step explanation:
Hi there!
Slope-intercept form: 
where m is the slope and b is the y-intercept (the value of y when the line crosses the y-axis)
<u>1) Determine the slope</u>
where the two given points are 
and 
Plug in the given points (-1, 4) and (0, 2)
Therefore, the slope of the line is -2. Plug this into :
<u>2) Determine the y-intercept</u>
Recall that the y-intercept is the value of y when the line crosses the y-axis, meaning that the y-intercept occurs when x is equal to 0.
One of the given points is (0,2). Notice how y=2 when x=0. Therefore, the y-intercept of the line is 2.
Plug this back into the equation:
I hope this helps!
<h3>
Answer: 23.13 square meters</h3>
===============================================================
Explanation:
The top and bottom semicircles can be joined to form a full circle. The same can be said about the left and right semicircles. Each circle has diameter 3 and radius 1.5
The area of one circle is
A = pi*r^2
A = 3.14*(1.5)^2
A = 7.065
So two circles doubles to 2*7.065 = 14.13 square meters in area.
Then the last step is to add on the area of the 3 by 3 square (area 3*3 = 9) to get 14.13+9 = 23.13
This area is approximate because pi = 3.14 is approximate. Use more decimal digits in pi to get a more accurate area. However, your teacher wants you to use this specific value so it's best to stick with 23.13
