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Elza [17]
3 years ago
11

Please help!!

Mathematics
1 answer:
morpeh [17]3 years ago
6 0

find from goggle please

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A plane intersects a triangular prism vertically as shown. Describe the cross-section shown here. A) pentagon B) rectangle C) tr
olga_2 [115]
The correct answer is c, but im not sure
3 0
3 years ago
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Explain why the expressions 8x(6-4) and 8x6-4 dont not have the same value even though they look very similar? HELP!! PLEASE EXP
viva [34]

Answer:

Because the first has () and the second doesn't so if u multiply the first one you will have to start with the () but if u answer the second u can start with anything

8 0
3 years ago
Is there a relationship between area or perimeter of a rectangle
Arturiano [62]
Hello,

For a given perimeter (P) there are an infinity of Area (A)
Let's say x the length, and y the wide of the rectangle

P=2(x+y)
A=xy
k=x-y >=0

As (x+y)²-4xy=(x-y)²: A²-4P=k² or P=(A²-k²)/4
In primus, you will find a graph (abacus) giving P for a A and k given.
Negative Area or P are excluded.(just remind the first quadrant, A>=0 and P>=0)

4 0
3 years ago
The present age of Sita's father is three times of Sita's present age. After 6 years the sum their age will be 69. Find their pr
Papessa [141]

Age of Sita is 14.25 years and her father is 42.75 years

Step-by-step explanation:

  • Step 1: From given details, let present age of Sita be x and that of her father be 3x.
  • Step 2: Form equations based on their ages in 6 years. (Given sum is 69)

⇒ (x + 6) + (3x + 6) = 69

⇒ 4x + 12 = 69

⇒ 4x = 57

⇒ x = 14.25

  • Step 3: Calculate age of Sita's father

⇒ 3x = 42.75

6 0
3 years ago
Solve the equation on the interval [0,2π). cos^4x=cos^4xcscx
Luba_88 [7]
\bf cos^4(x)=cos^4(x)csc(x)\\\\
-----------------------------\\\\
cos^4(x)=cos^4(x)\cfrac{1}{sin(x)}\implies cos^4(x)=\cfrac{cos^4(x)}{sin(x)}
\\\\\\
cos^4(x)sin(x)=cos^4(x)\implies cos^4(x)sin(x)-cos^4(x)=0
\\\\\\
cos^4(x)[sin(x)-1]=0\to 
\begin{cases}
cos^4(x)=0\to x=cos^{-1}(0)\\
----------\\
sin(x)-1=0\\
sin(x)=1\to x=sin^{-1}(1)
\end{cases}

\bf \measuredangle x = cos^{-1}(0)\implies \measuredangle x = 
\begin{cases}
\frac{\pi }{2}\\\\
\frac{3\pi }{2}
\end{cases}
\\\\\\
\measuredangle x=sin^{-1}(1)\implies \measuredangle x=\frac{\pi }{2}

8 0
4 years ago
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