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kotykmax [81]
3 years ago
6

A middle school has the fifth and sixth grades. There are 100 fifth-grade boys and 110 fifth-grade girls. There are 7 fewer sixt

h-grade boys than fifth-grade boys, and there are 10 more sixth-grade girls than sixth-grade boys What is the ratio of girls to boys in the middle school? Show your work.
Mathematics
1 answer:
Mama L [17]3 years ago
4 0

The ratio of girls to boys in the middle school = 213 : 193

Number of fifth-grade boys = 100

Number of fifth grade girls = 110

There are 7 fewer sixth-grade boys than fifth-grade boys

Number of sixth-grade boys = 100 - 7

Number of sixth-grade boys = 93

There are 10 more sixth-grade girls than sixth-grade boys

Number of sixth-grade girls = 93 + 10

Number of sixth-grade girls = 103

Total number of girls = Number of sixth-grade girls + Number of fifth-grade girls

Total number of girls = `103 + 110

Total number of girls = 213

Total number of boys = Number of sixth-grade boys + Number of fifth-grade boys

Total number of boys = `93 + 100

Total number of boys = 193

The ratio of girls to boys in the middle school = 213 : 193

Learn more on ratio here: brainly.com/question/25927869

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Evaluate the line integral by the two following methods. xy dx + x2y3 dy C is counterclockwise around the triangle with vertices
nadezda [96]

Answer:

a)

\frac{2}{3}

b)

\frac{2}{3}

Step-by-step explanation:

a) The first part requires that we use line integral to evaluate directly.

The line integral is

\int_C xydx +  {x}^{2}  {y}^{3} dy

where C is counterclockwise around the triangle with vertices (0, 0), (1, 0), and (1, 2)

The boundary of integration is shown in the attachment.

Our first line integral is

L_1 = \int_ {(0,0)}^{(1,0)} xydx +  {x}^{2}  {y}^{3} dy

The equation of this line is y=0, x varies from 0 to 1.

When we substitute y=0 every becomes zero.

\therefore \: L_1 =0

Our second line integral is

L_2 = \int_ {(1,0)}^{(1,2)} xydx +  {x}^{2}  {y}^{3} dy

The equation of this line is:

x = 0 \implies \: dx = 0

y varies from 1 to 2.

We substitute the boundary and the values to get:

L_2 = \int_ {1}^{2}1 \cdot y(0) +  {1}^{2}   \cdot \: {y}^{3} dy

L_2 = \int_ {1}^2 {y}^{3} dy =  \frac{8}{3}

The 3rd line integral is:

L_3 = \int_ {(1,2)}^{(0,0)} xydx +  {x}^{2}  {y}^{3} dy

The equation of this line is

y = 2x \implies \: dy = 2dx

x varies from 0 to 1.

We substitute to get:

L_3 = \int_ {1}^{0} x \cdot \: 2xdx +  {x}^{2}  {(2x)}^{3}(2 dx)

L_3 = \int_ {1}^{0} 8 {x}^{5}  + 2 {x}^{2} dx  =  - 2

The value of the line integral is

L = L_1 + L_2 + L_3

L = 0 +  \frac{8}{3}  +  - 2 =  \frac{2}{3}

b) The second part requires the use of Green's Theorem to evaluate:

\int_C xydx +  {x}^{2}  {y}^{3} dy

Since C is a closed curve with counterclockwise orientation, we can apply the Green's Theorem.

This is given by:

\int_C \: Pdx +Q  \: dy =  \int \int_ R \: Q_y -  P_x \: dA

\int_C \: xydx + {x}^{2} {y}^{3}   \: dy =  \int \int_ R \: 3 {x}^{2}  {y}^{2}  -  y \: dA

We choose our region of integration parallel to the y-axis.

\int_C \: xydx + {x}^{2} {y}^{3}   \: dy =  \int_ 0^{1} \int_ 0^{2x}  \: 3 {x}^{2}  {y}^{2}  -  y \: dydx

\int_C \: xydx + {x}^{2} {y}^{3}   \: dy =  \int_ 0^{1} \:  {x}^{2}  {y}^{3}  -   \frac{1}{2}  {y}^{2} |_ 0^{2x}  dx

\int_C \: xydx + {x}^{2} {y}^{3}   \: dy =  \int_ 0^{1} \:  8{x}^{5} -  2 {x}^{2}   dx =  \frac{2}{3}

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