So you wanna find the area of each surface:
For the large square surface in the top you do 12*15=180 and there are two large squares faces so you multiply 180 by 2 which equals 360 then you find the small rectangle faces: 5*15=75 and multiply by 2 Bc there are 2 faces like that so it would be 75*2=150 then for the last faces you do 12*5= 60 *2= 120 at the end I add all of the faces up, 360+150+120=630
Answer:
red
Step-by-step explanation:
1-red
2-blue
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5-red
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23-red
im so smart fr
53.73 rounded to the nearest whole number would be C) 54.
***hope this helps:) brainliest? ;)
Answer:
325
Step-by-step explanation:
You must have heard about Arithmetic Progressions (AP)
Arithmetic progressions are a series of numbers such that every successive number is the sum of a constant number and the previous number.
Our very own counting numbers form AP
For example :-
2 = 1 + <u>1</u>
3 = 2 + <u>1</u>
4 = 3 + <u>1</u>
The number in bold (1) is that constant number which is added to a number to form its successive number.
To find the sum of series forming AP, we use the formula :-
here,
- n is the number of terms
- a is the first number of the series
- an is the last number of the series
So we'll use all this information to find the sum of continuous numbers from 1 to 25 where 1 is the first term(a) and 25 is the last(an).
and n is 25
So, the value of S comes out to be 325.
Substitute the given point (3, -2) in each condition
(i) y < -3; y ≤ (2/3)x - 4
-2 < -3 → False
Therefore, it is not a solution.
[Since the values of x and y must satisfy both the conditions, if one of them does not satisfy the condition then it is not necessary to check the other one.]
(ii) y > -3; y ≥ 2/3x - 4
-2 > -3 → True
y ≥ 2/3x - 4
-2 ≥ -2 → True
Therefore, it is a solution.
(iii) y < -3; y ≥ 2/3x - 4
-2 < -3 → False
Therefore, it is not a solution.
(iv) y > -2; y ≤ 2/3x - 4
-2 > -2 → False
Therefore, it is not a solution.
Thereofore, the system of linear inequalities having the point (3, -2) in its solution set is y > -3; y ≥ 2/3x - 4. Hope it helps you.
