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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Answer:
Please check the explanation.
Step-by-step explanation:
Finding Domain:
We know that the domain of a function is the set of input or argument values for which the function is real and defined.
From the given graph, it is clear that the starting x-value of the line is x=-2, the closed circle at the starting value of x= -2 means the x-value x=-2 is included.
And the line ends at the x-value x=1 with a closed circle, meaning the ending value of x=1 is also included.
Thus, the domain is:
D: {-2, -1, 0, 1} or D: −2 ≤ x ≤ 1
Finding Range:
We also know that the range of a function is the set of values of the dependent variable for which a function is defined
From the given graph, it is clear that the starting y-value of the line is y=0, the closed circle at the starting value of y = 0 means the y-value y=0 is included.
And the line ends at the y-value y=2 with a closed circle, meaning the ending value of y=2 is also included.
Thus, the range is:
R: {0, 1, 2} or R: 0 ≤ y ≤ 2
Answer:
D. 65°
Step-by-step explanation:
it's a isosceles triangle so the angles of same length sides has same measure.
The rental costs $0.25 per 1 mile
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Answer:
95.1CM²
Step-by-step explanation:
THERE ARE 3 SHAPES
-AREA OF RECTANGLE=LXB
=21 X 3 = 63 CM²
-AREA OF SQUARE = SXS
=5 X 5= 25CM²
-AREA OF CIRCLE = Pi x r²
RADIUS = HALF OF DIAMETER = 1.5
=3.14 X 1.5 ²
=3.14 X 2.25
=7.065
TOTAL AREA= 63 + 25 + 7.065=
ROUNDED = OPTION B
=95.1 CM²
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