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3 years ago

Find the area of a playing field whichs area is 15 meters

Mathematics

1 answer:

uranmaximum [27]3 years ago

7 0

Easy just multiply 4 and 15 bc a playing field has 4 sides and u get 60

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Find the area of a trapezoid with b1 = 3 cm, b2 = 22 cm, and h = 6 cm.

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Area of trapezoid = Average length of parallel lines x height

3+22=25
25/2= 12.5
12.5x6=75cm^2

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3 years ago

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The table represents the start of the division of 15x3 + x2 - 3x + 2 by 3x + 2.

9966 [12]

The result of dividing $15x^{3}+x^{2}-3x+2$ by $3x+2$ is $5 x^{2} -3x+1$ with no remainder. Please check the step by step procedures in the picture attached.
We can conclude that:
The quotient of the division is $5 x^{2} -3x+1$
The remainder of the division is $0$
The divisor is $3x+2$
The dividend is $15x^{3}+x^{2}-3x+2$

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4 years ago

Hii pls help i’ll give brainliest if you give a correct answer

iragen [17]

Hey there ..

Ur answer .. Check it out ..

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3 years ago

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aleksandr82 [10.1K]

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3 years ago

Please help with this Calculus questions

Triss [41]

Answer:

$\int_{0}^{1}\left( - \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2 \right)du=\frac{193}{100}=1.93$ .

Step-by-step explanation:

To find $\int_{0}^{1}\left( - \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2 \right)du$ .

First, calculate the corresponding indefinite integral:

Integrate term by term:

$\int{\left(- \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2\right)d u}} =\int{2 d u} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u}$

Apply the constant rule $\int c\, du = c u$

$\int{2 d u}} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u} = {\left(2 u\right)} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u}$

Apply the constant multiple rule $\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$

$2 u - {\int{\frac{3 u^{9}}{2} d u}} + \int{\frac{2 u^{4}}{5} d u} = 2 u - {\left(\frac{3}{2} \int{u^{9} d u}\right)} + \left(\frac{2}{5} \int{u^{4} d u}\right)$

Apply the power rule $\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$

$2 u - \frac{3}{2} {\int{u^{9} d u}} + \frac{2}{5} {\int{u^{4} d u}}=2 u - \frac{3}{2} {\frac{u^{1 + 9}}{1 + 9}}+ \frac{2}{5}{\frac{u^{1 + 4}}{1 + 4}}$

Therefore,

$\int{\left(- \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2\right)d u} = - \frac{3 u^{10}}{20} + \frac{2 u^{5}}{25} + 2 u = \frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)$

According to the Fundamental Theorem of Calculus, $\int_a^b F(x) dx=f(b)-f(a)$ , so just evaluate the integral at the endpoints, and that's the answer.

$\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=1\right)}=\frac{193}{100}$

$\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=0\right)}=0$

$\int_{0}^{1}\left( - \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2 \right)du=\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=1\right)}-\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=0\right)}=\frac{193}{100}$

6 0

3 years ago