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Aliun [14]
4 years ago
11

what is the area in square centimeters of the trapezoid below the height is 6.4 and the base is 12.9 and 8.6

Mathematics
2 answers:
Leona [35]4 years ago
4 0
Hello!
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The area in square centimeters is 68.8.
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Have a great day!
Lelu [443]4 years ago
3 0

Answer:

The area of given trapezoid is 68.8 cm²

Step-by-step explanation:

Given: A trapezoid with height 6.4 cm and parallel base is 12.9 cm and 8.6 cm.

We have to find the area of area of this trapezoid.

Area of trapezoid = \frac{1}{2}\times \text{sum of parallel sides}\times height

Given : Height = 6.4 cm

and parallel base is 12.9 cm and 8.6 cm.

Substitute, we have,

Area of trapezoid = \frac{1}{2}\times (12.9+8.6)\times 6.4

Simplify, we have,

Area of trapezoid = \frac{1}{2}\times 21.5\times 6.4

Area of trapezoid = 68.8 cm²

Thus, The area of given trapezoid is 68.8 cm²

           

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The Lagrangian is

L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)

Take your partial derivatives and set them equal to 0:

\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}

Adding the first three equations together yields

2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43

and plugging this into the first three equations, you find a critical point at (x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right).

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The sum of two terms of gp is 6 and that of first four terms is 15/2.Find the sum of first six terms.​
Gnoma [55]

Given:

The sum of two terms of GP is 6 and that of first four terms is \dfrac{15}{2}.

To find:

The sum of first six terms.​

Solution:

We have,

S_2=6

S_4=\dfrac{15}{2}

Sum of first n terms of a GP is

S_n=\dfrac{a(1-r^n)}{1-r}              ...(i)

Putting n=2, we get

S_2=\dfrac{a(1-r^2)}{1-r}

6=\dfrac{a(1-r)(1+r)}{1-r}

6=a(1+r)                    ...(ii)

Putting n=4, we get

S_4=\dfrac{a(1-r^4)}{1-r}

\dfrac{15}{2}=\dfrac{a(1-r^2)(1+r^2)}{1-r}

\dfrac{15}{2}=\dfrac{a(1+r)(1-r)(1+r^2)}{1-r}

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Divide both sides by 6.

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\dfrac{5-4}{4}=r^2

\dfrac{1}{4}=r^2

Taking square root on both sides, we get

\pm \sqrt{\dfrac{1}{4}}=r

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6=a(1+0.5)  

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S_6=\dfrac{4(1-(0.5)^6)}{(1-0.5)}

S_6=\dfrac{4(1-0.015625)}{0.5}

S_6=8(0.984375)

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Case 2: If r is negative, then using (ii) we get

6=a(1-0.5)  

6=a(0.5)  

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S_6=\dfrac{12(1-0.015625)}{1.5}

S_6=8(0.984375)

S_6=7.875

Therefore, the sum of the first six terms is 7.875.

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