What's the surface area? (EXPLAIN YOUR ANSWER)

Dimas [21]

Simplify $\frac{2}{5}$ (d - 10) to $\frac{2(d - 10)}{5}$ :
\frac{2(d - 10)}{5} [/tex] - 23(d + 6) x 25(d - 10) - 23(d + 6)

Simplify 23(d + 6) x 25(d - 10) to 575(d + 6)(d - 10) :
\frac{2(d - 10)}{5} [/tex] - 575(d + 6)(d - 10) - 23(d + 6)

Then Expand :
\frac{2(d - 10)}{5} [/tex] - 575 $d^{2}$ + 5750d - 3450d + 34500 - 23d - 138

Now Collect Like Terms :
\frac{2(d - 10)}{5} [/tex] - 575 $d^{2}$ + (5750d - 3450d + 34500 - 23d) + (34500 - 138)

Answer :
\frac{2(d - 10)}{5} [/tex] - 575 $d^{2}$ + 2277d + 34362

7 0

3 years ago

What is 6.25 as a fraction and percent

ipn [44]

Convert the decimal by multiplying by 100.625%

Convert to a fraction by placing the decimal number over a power of 10.254

5 0

3 years ago

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What two numbers multiply to -36 and add up to -9

Softa [21]

Answer:

-12,3

Step-by-step explanation:

-12×3=-36

-12+3=-9

3 0

3 years ago

In evaluating a double integral over a region D, a sum of iterated integrals was obtained as follows:

BabaBlast [244]

Answer

$a=0$ , $b=2$

$g_1(x)=\frac{5x}{2}$ , $g_2(x)=7-x$

Step-by-step explanation:

Given that

$\int \int Df(x,y)dA=\int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy+\int_5^7\int_0^{7-y} f(x,y)dxdy\; \cdots (i)$

For the term $\int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy$ .

Limits for $x$ is from $x=0$ to $x=\frac {2y}{5}$ and for $y$ is from $y=0$ to $y=5$ and the region $D$ , for this double integration is the shaded region as shown in graph 1.

Now, reverse the order of integration, first integrate with respect to $y$ then with respect to $x$ . So, the limits of $y$ become from $y=\frac{5x}{2}$ to $y=5$ and limits of $x$ become from $x=0$ to $x=2$ as shown in graph 2.

So, on reversing the order of integration, this double integration can be written as

$\int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy=\int_0 ^2\int _ {\frac {5x}{2}}^5 f(x,y)dydx\; \cdots (ii)$

Similarly, for the other term $\int_5 ^7\int _0 ^ {7-y} f(x,y)dxdy$ .

Limits for $x$ is from $x=0$ to $x=7-y$ and limits for $y$ is from $y=5$ to $y=7$ and the region $D$ , for this double integration is the shaded region as shown in graph 3.

Now, reverse the order of integration, first integrate with respect to $y$ then with respect to $x$ . So, the limits of $y$ become from $y=5$ to $y=7-x$ and limits of $x$ become from $x=0$ to $x=2$ as shown in graph 4.

So, on reversing the order of integration, this double integration can be written as

$\int_5 ^7\int _0 ^ {7-y} f(x,y)dxdy=\int_0 ^2\int _5 ^ {7-x} f(x,y)dydx\;\cdots (iii)$

Hence, from equations $(i)$ , $(ii)$ and $(iii)$ , on reversing the order of integration, the required expression is

$\int \int Df(x,y)dA=\int_0 ^2\int _ {\frac {5x}{2}}^5 f(x,y)dydx+\int_0 ^2\int _5 ^ {7-x} f(x,y)dydx$

$\Rightarrow \int \int Df(x,y)dA=\int_0 ^2\left(\int _ {\frac {5x}{2}}^5 f(x,y)+\int _5 ^ {7-x} f(x,y)\right)dydx$

$\Rightarrow \int \int Df(x,y)dA=\int_0 ^2\int _ {\frac {5x}{2}}^{7-x} f(x,y)dydx\; \cdots (iv)$

Now, compare the RHS of the equation $(iv)$ with

$\int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)dydx$

We have,

$a=0, b=2, g_1(x)=\frac{5x}{2}$ and $g_2(x)=7-x$ .

3 0

3 years ago

I need help on this question!!!!

blsea [12.9K]

D
just subtract 14 from 26

6 0

3 years ago

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What's the surface area?(EXPLAIN YOUR ANSWER)