Answer:
24s^2, 54s^2, 96s^2
Step-by-step explanation:
Let s represent the initial side length of the cube. Then the area of each face of the cube is A = 6s^2 (recalling that the area of a square of side length s is s^2).
a) Now suppose we double the side length. The total area of the 6 faces of the cube will now be A = 6(2s)^2, or 24s^2 (a 24 times larger surface area),
b) tripled: A = 6(3s)^2 = 54x^2
c) quadrupled? A = 6(4s)^2 = 96s^2
Vertical angles are congruent...they are equal....so set them equal to each other and solve for x
6x - 22 = 4x + 2
6x - 4x = 2 + 22
2x = 24
x = 24/2
x = 12 <==
<span>If this is an isosceles triangle, then it has two 45 degree angles corresponding to two legs of equal length. Orient the base of this triangle so that it's horizontal, and represent its length by b. Let h represent the height of the triangle. Then the area of this right triangle is 50 square inches = (1/2)(b)(h), or A = (b/2)h = 50 in^2.
Due to the 45 degree angles, the height of this triangle is equal to half the base, or h = b/2. Thus, (b/2)h = 50 becomes (b/2)(b/2) = 50, or b^2=200. Thus, b = 10sqrt(2), and h=(1/2)(10 sqrt(2)), or h = 5sqrt(2).
The length of one of the legs is the sqrt of [5sqrt(2)]^2+[5sqrt(2)]^2, or
sqrt(25(2)+25(2)) = sqrt(100) = 10.
</span>
Answer:
x = 5
Step-by-step explanation:
Given that,
NL = 5x, LM = 3x, and NL = 15
L is between M and N.
To find,
The value of x.
Solution,
It is mentioned that, L is between M and N. So,
LM = LN
Put all the values,
3x = 15
⇒ x = 5
So, the value of x is equal to 5.
Reference,
brainly.com/question/19871240
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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