All categories

3 years ago

A line that includes the points (9,j) and (10,4) has a slope of 1. What is the value of j?

Mathematics

2 answers:

Sergeu [11.5K]3 years ago

5 0

Answer:

soln

A=(9,j)

x1=9,y1=j

B=(10,4)

x2=10,y2=4

slope=1

by using the fomulaof slope,

slope=y2-y1/x2-x1

1 =4-j/10-9

1 = 4-j/1

1=4-j

j=4-1

j=3

NNADVOKAT [17]3 years ago

5 0

Answer:

j = 3

Step-by-step explanation:

(9 , j) ; $x_{1}= 9$ and $y_{1} = j$

(10, 4) ; $x_{2} = 10$ and $y_{2} = 4\\\\$

$Slope = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\ \frac{4 - j}{10-9} = 1\\\\\frac{4 - j}{1}=1$

4 - j = 1 {Subtract 4 from both sides}

-j = 1 -4

-j = -3 {Multiply both sides by (-1) }

j = 3

You might be interested in

A report by the US Geological Survey indicates that glaciers in Glacier National Park, Montana, are shrinking. Recent estimates

postnew [5]

Answer

a) C

b) B

c) C, E

d) 15.5

e) 0.7

f) 67.7

Step-by-step explanation

a) A linear equation can be written as $y=mx+c$ where $m$ is the slope and $c$ is the intercept on the $y$ -axis. The slope describes how $y$ changes when $x$ changes. A negative value means the it is a decreasing change. In our case, $y$ is $A$ and $x$ is $t$ . Thus, the slope of the equation is $m=-0.062$ . This means that the area of glacier is decreasing by $0.062 \text{km}{}^2 = 62000 \text{m}{}^2$ every year since 2000.

b) The $A$ -intercept (= 16.2 $\text{km}{}^2$ ) represents the area covered when $t=0$ . But $t=0$ starting from year 2000. This intercept then represents the area covered by glacier in the year 2000. Therefore, the area covered by glacier in 2000 was 16.2 $\text{km}{}^2$ .

c) $A=f(t)$ means the area covered after year 2000. Setting it to 12 means the area covered after $t$ years is 12 $\text{km}{}^2$ ). $t$ is therefore the number of years since 2000 that the total area covered will be 12 $\text{km}{}^2$ ).

d) $f(12)=16.2−0.062\times12 =16.2-0.744 = =15.456 15.5 \text{km}{}^2$ to 1 decimal place.

e) The term involving $t$ represents the disappearing area. In 12 years, the disappeared area is $0.062\times12=0.744 =0.7 \text{km}{}^2$ to 1 decimal place.