Now that we’ve learned how to solve word problems involving the sum of consecutive integers, let’s narrow it down and this time, focus on word problems that only involve finding the sum of consecutive even integers.
But before we start delving into word problems, it’s important that we have a good understanding of what even integers, as well as consecutive even integers, are.
Even Integers
We know that even numbers are integers that can be divided exactly or evenly by 22. Thus, the general form of the even integer nn, is n = 2kn=2k, where kk is also an integer.
In other words, since even numbers are the multiples of 22, we can represent an even integer nn by 2k2k, where kk is also an integer. So if we have the even integers 1010 and 1616,
Answer:
1 is 136, 2 is 44, annd 4 is 44
Step-by-step explanation:
To find the surface area of the open net, you’ll have to find the area of the faces
So, since there are 6 boxes you’ll have to find the area of the faces and add it up to find the surface area
10x4=40x4=160
3x4=12x2=24
160+24=184
Hope this helps ;)
Answer:
D
Step-by-step explanation:
Answer:
titutex=cos\alp,\alp∈[0:;π]
\displaystyle Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1)\Leftright |cos\alp +sin\alp |=\sqrt{2}(2cos^2\alp -1)Then∣x+
1−x
2
∣=
2
(2x
2
−1)\Leftright∣cos\alp+sin\alp∣=
2
(2cos
2
\alp−1)
\displaystyle |\N {\sqrt{2}}cos(\alp-\frac{\pi}{4})|=\N {\sqrt{2}}cos(2\alp )\Right \alp\in[0\: ;\: \frac{\pi}{4}]\cup [\frac{3\pi}{4}\: ;\: \pi]∣N
2
cos(\alp−
4
π
)∣=N
2
cos(2\alp)\Right\alp∈[0;
4
π
]∪[
4
3π
;π]
1) \displaystyle \alp \in [0\: ;\: \frac{\pi}{4}]\alp∈[0;
4
π
]
\displaystyle cos(\alp -\frac{\pi}{4})=cos(2\alp )\dotscos(\alp−
4
π
)=cos(2\alp)…
2. \displaystyle \alp\in [\frac{3\pi}{4}\: ;\: \pi]\alp∈[
4
3π
;π]
\displaystyle -cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots−cos(\alp−
4
π
)=cos(2\alp)…
1
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