Please help me find the area

Mathematics

2 answers:

photoshop1234 [79]2 years ago

8 0

Answer:

240 square units

Step-by-step explanation:

rectangle= 20x10=200

triangle= 1/2(4)(20)=40

200+40=240

slavikrds [6]2 years ago

4 0

Answer:

240 units squared

Step-by-step explanation:

The square is 20 by 10 so 20x10=200. To do the area of a triangle you take the length of the bottom of the shape (20) and the height (4) multiply them and then divide by 2. So 20x4=80 divided by 2 = 40. Then you add the two numbers together to get the full area 200+40=240 units squared.

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Evaluating more integrals

Greeley [361]

(a) The integral is equal to the area of the triangle; it has height 20 and base 10, so the area is 20*10/2 = 100.

(b) The integral is equal to the area of the semicircle with radius 10. It's also under the horizontal axis, so the area is negative. The semicircle has area $\frac{\pi10^2}2=50\pi$ , so the integral is -50π.

$\displaystyle\int_{30}^{35}g(x)\,\mathrm dx$

which is the area of the triangle on the right. It has height and base 5, so its area is 25/2.

Then split up the desired integral as

$\displaystyle\int_0^{35}g(x)\,\mathrm dx=\int_0^{10}g(x)\,\mathrm dx+\int_{10}^{30}g(x)\,\mathrm dx+\int_{30}^{35}g(x)\,\mathrm dx$

and plug in the integral values you know:

$\displaystyle\int_0^{35}g(x)\,\mathrm dx=100-50\pi+\frac{25}2=\frac{225}2-50\pi$

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For the polynomial function ƒ(x) = x4 − 12x3 + 46x2 − 60x + 25, find all local and global extrema.

rodikova [14]

$\bf f(x)=x^4-12x^3+46x^2-60x+25\\\\
-------------------------------\\\\
\cfrac{df}{dx}=4x^3-36x^2+92x-60\implies 0=4x^3-36x^2+92x-60
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0=x^3-9x^2+23x-15\impliedby \textit{doing some synthetic division with 5}
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0=(x^2-4x+3)(x-5)\implies 0=(x-3)(x-1)(x-5)\\\\
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\begin{cases}
f(1)=0\leftarrow minimum&1,0\\
f(3)=16\leftarrow maximum&3,16\\
f(5)=0\leftarrow minimum&5,0
\end{cases}\impliedby \textit{3 extrema, locals and globals}$