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3 years ago

12

a team of 4 people participate in a 400-yard relay race each team member runs the same distance the team completes the race in 5

3.2 seconds what is the average running time for each person

1 answer:

Korvikt [17]3 years ago

6 0

13.3s you take 53.2 and divide it by the total number of people making it an average of 13.3 per 100m

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Juan is making t-shirts to sell at a concert.

o-na [289]

Answer:

7 shirts

Step-by-step explanation:

The function for producing shirts is

f($) = 50 + 7.5(x), where x is the number of shirt

The function for selling shirts is

f($) = 15(x)

For there to be a profit, he must make more money selling the shirts than he spent on making the shirts

Therefore, 15(x) > 50 + 7.5(x)

15x > 50 + 7.5x

15x - 7.5x > 50

7.5x > 50

x > 6.67 shirts

To make a profit, Juan must sell at least 7 shirts

7 0

2 years ago

2) 3x + 4y = 8<br> 2x+y=42<br> a) (30, -19)<br> b) (31, -20)<br> c) (32,-22)<br> What is the answer

kirza4 [7]

Answer:

C

Step-by-step explanation:

We can solve simultaneous equations using substitution method, elimination method or graphical method. But for this purpose, we will be using the elimination method.

3x+4y=8 Equation 1

2x+y=42 Equation 2

Multiply Equation 1 by 2 and equation 2 by 3, so as to get the same coefficient for x

2(3x+4y=8)= 6x+8y=16 Equation 3

3(2x+y=42)= 6x+3y=126 Equation 4

Subtract equation 4 from 3, to eliminate x

6x-6x=0

8y-3y= 5y

16-126= -110

We now have 5y=-110

Divide both sides by 5,

y= -110/5

= -22

Substituting for y in equation 2

2x+(-22)= 42

2x= 42+22

2x=64

x= 64/2

= 32

(x, y)

(32, -22)

7 0

3 years ago

The given line segment has a midpoint at (3, 1).

Katena32 [7]

Answer:

$y=\frac{1}{3}x$

Step-by-step explanation:

The given line segment has a midpoint at (3, 1) and goes through (2, 4), (3, 1), and (4, -2). We can use any two of the three points to calculate the equation of the line. Let us use the points (2, 4) and (4, -2)

Therefore the line goes through (2, 4) and (4, -2). The equation of a line passing through $(x_1,y_1)\ and\ (x_2,y_2)$ is:

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$ .

Therefore the line passing through (2, 4) and (4, -2) has an equation:

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}\\\frac{y-4}{x-2}=\frac{-2-4}{4-2}\\\frac{y-4}{x-2}=\frac{-6}{2}\\y-4=x-2(-3)\\y-4=-3x+6\\y=-3x+10$

Comparing with the general equation of line: y = mx + c, the slope (m) = -3 and the intercept on the y axis (c) = 10

Two lines are said to be perpendicular if the product of their slope is -1. If the slope of line one is m1 and the slope of line 2 = m2, then the two lines are perpendicular if:

$m_1m_2=-1$ .

Therefore The slope (m2) of the perpendicular bisector of y = -3x + 10 is:

$m_1m_2=-1\\-3m_2=-1\\m_2=\frac{1}{3}$

Since it is the perpendicular bisector of the given line segment, it passes through the midpoint (3, 1). The equation of the perpendicular bisector is:

$\frac{y-y_1}{x-x_1}=m\\\frac{y-1}{x-3}=\frac{1}{3}\\ y-1= \frac{1}{3}(x-3)\\ y-1=\frac{1}{3}x-1\\y=\frac{1}{3}x$

the equation, in slope-intercept form, of the perpendicular bisector of the given line segment is $y=\frac{1}{3}x$

7 0

3 years ago

Read 2 more answers

What are the zeros of the polynomial function y=2x^3-7x^2+2x+3?

zlopas [31]

$y = 2 {x}^{3} - 7 {x}^{2} + 2x + 3$

By inspection 1 is a root of y ,
Hence , (x-1) is a factor of y ,

$\dfrac{2 {x}^{3} - 7 {x}^{2} + 2x + 3 }{x - 1} = 2 {x}^{2} - 5x - 3 \\ \\ 2 {x}^{2} - 5x - 3 = (2x + 1)(x - 3)$

Hence , the 2 other roots are -1/2 and 3 , and the above graphs confirms the same.

Hope it helps you :)

7 0

3 years ago

Please help me with both questions!!!!

Naya [18.7K]

Answer:

Step-by-step explanation:

16.

points are (0,-3),(2,1),(4,-3)

eq. of line through (0,-3) and (2,1) is

y+3=\frac{1+3}{2-0}(x-0)

or y+3=2x

2x-y=3

as the line is dotted so either < or >.

consider 2x-y>3

put x=0,y=0

0>3

which is not possible .

(0,0) does not satisfy the inequality

Hence shaded region is the required region.

now eq of line through (2,1)and (4,-3) is

y-1=\frac{-3-1}{4-2}(x-2)

y-1=-2(x-2)

y-1=-2x+4

2x+y=5

it is also a dotted line so either < or >

consider 2x+y<5

put x=0,y=0

0<5

which is true.

so (0,0) satisfy this inequality.

so the two inequalities are

2x-y>3

and 2x+y<5

17.

consider the points (0,4),(2,3),(4,4)

eq. of line through (0,4) and (2,3) is

y-4=\frac{3-4}{2-0}(x-0)

y-4=-1/2(x)

2y-8=-x

or x+2y=8

as the line is solid

so either≤ or ≥

consider x+2y≥8

put x=0,y=0

0≥8

which is impossible.

(0,0) does not satisfy the graph.

which is true as graph lies above the line.

again eq. of line through (2,3) and (4,4) is

y-3=\frac{4-3}{4-2}(x-2)

y-3=1/2(x-2)

2y-6=x-2

x-2y=-4

consider x-2y≤-4

put x=0, y=0

0≤-4

which is impossible.

so (0,0) does not satisfy the graph.

so inequality is true as shaded portion is above and left of the line.

so two inequalities are

x+2y≥ 8

and x-2y≤-4

5 0

3 years ago