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Mathematics

2 answers:

USPshnik [31]3 years ago

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Dvinal [7]3 years ago

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The sum of two numbers is 54.The smaller number is 22 less than the larger number.what are the numbers?

lukranit [14]

Answer:

x=16 y=38

Step-by-step explanation:

x+y=54.......equation 1

x=y-22........ equation 2

substitute

y-22+y=54

2y-22=54

2y=54+22

2y=76

y=76/2

y=38''

x=y-22

x=38-22

x=16''

8 0

3 years ago

Is (4, 8) a solution of y > 5x + 4?

lana66690 [7]

This is not a solution. When plugged in,
8 > 20 + 4.
8 > 24, which is not true. :)

8 0

3 years ago

A band sells tickets to a concert cost either $15 for adults or $10 for students. A total of 150 tickets are sold. If the band m

kodGreya [7K]

Answer:

Number of adult tickets sold= 100

Step-by-step explanation:

Giving the following information:

Adults tickets= $15

Student tickets= $10

Number of tickets sold= 150

Total sales= $2,000

First, we determine the systems of equations:

15*x + 10*y= 2,000

x + y = 150

x= number of adults tickets sold

y= number of students tickets sold

Now, we isolate x in one equation, and substitute it in the other one:

x= 150 - y

15*(150 - y) + 10y = 2,000

2,250 - 15y + 10y = 2,000

250 = 5y

50= y

x= 150 - 50

x= 100

Prove:

15*100 + 10*50= 2,000

100 + 50 = 150

3 0

3 years ago

Read 2 more answers

The graph of a quadratic function with vertex (3,-2) is shown in the figure below.

fgiga [73]

Answer:

Domain: all real numbers

Range: less than/equal to -2

8 0

3 years ago

Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (1, 5), F (3, 5), a

hram777 [196]

Answer:

$(1,1)$

Step-by-step explanation:

Given: E, F, G, H denote the three coordinates of the area fenced

To find: coordinates of point H

Solution:

According to distance formula,

length of side joining points $(x_1,y_1)\,,\,(x_2,y_2)$ is equal to $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

So,

$EF=\sqrt{(3-1)^2+(5-5)^2}=2\,\,units\\FG=\sqrt{(6-3)^2+(1-5)^2}=\sqrt{9+16}=5\,\,units\\GH=\sqrt{(x-6)^2+(y-1)^2}\\EH=\sqrt{(x-1)^2+(y-5)^2}$

Perimeter of a figure is the length of its outline.

$EF+FG+GH+EH=16\\2+5+\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16-2-5\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$

Put $(x,y)=(1,1)$

$\sqrt{(1-6)^2+(1-1)^2}+\sqrt{(1-1)^2+(1-5)^2}=9\\\sqrt{25}+\sqrt{16}=9\\5+4=9\\9=9$

This is true.

So, the point $(1,1)$ satisfies the equation $\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$

So, point H is $(1,1)$ .

7 0

3 years ago