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Paladinen [302]
3 years ago
11

Ms.Lamberson got a new puppy abd weighed it in September and October in September the Puppy Weighed9 and one Fourth pounds In Oc

tober the Puppy Weighed 11and One Half Pounds how much heavier was the puppy in October than it was in September
Mathematics
1 answer:
ddd [48]3 years ago
8 0

Answer: The puppy was 2.25\ lb havier in October that it was in September.

Step-by-step explanation:

You need to make the conversion from mixed numbers to decimal numbers. To do it, divide the numerator of the fraction by the denominator add the result to whole part number:

9\frac{1}{4}=9+0.25=9.25\\\\11\frac{1}{2}=11+0.5=11.5

The weight of the puppy in September was:

w_1=9.25\ lb

And its weight in October was:

w_2=11.5\ lb

Subtract its weight in September from its weight in October in order to find how much heavier the puppy was in October than it was in September:

w_2-w_1=11.5\ lb-9.25\ lb=2.25\ lb

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4 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
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aleksley [76]

N = 3

Step-by-step explanation: If you do the equation it comes out to n=4

n+3n=12

4n=12

n=3

You have to add the like terms then divide it so the variable is by itself. And whatever you do to one side, you have to do to the other

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Answer: 160

Step-by-step explanation:

160 is the answer trust me

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