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4 years ago

What is the linear function equation represented by the graph?

Mathematics

1 answer:

Tom [10]4 years ago

8 0

Answer:

Y= -1/2x + 1

Step-by-step explanation:

The equation for a line is Y=Mx+B

M stands for the Slope.

-Slope is rise/run. So, for this graph the line "rises" 1 point and "runs" left 2 points until meeting the line again. This slope is negative because it "runs" to the left.

B stands for the Y-Intercept

-The Y-intercept is the point at which the line crosses the Y axis. For this example, the line crosses the Y-intercept at (0,1).

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4.66666666667 as a fraction. (They all put 466666666667/100000000000)

natali 33 [55]

<h2>Answer:</h2><h2><em><u>Brainliest plz</u></em></h2>

Step-by-step explanation:

that is because you have to write it like this:

4.666...

in fractions its 4 and 2/3

$4 \frac{2}{3}$

7 0

3 years ago

Read 2 more answers

What’s the correct answer for this ?

icang [17]

Answer:

Step-by-step explanation:

Since both the triangles as kept separately are similar, so we'll take proptionality of their sides to find one side

MP/ML=MN/MK

20/28=35/MK

CROSS MULTIPLYING

20×MK=28×35

MK=980/20

MK=49

5 0

3 years ago

Here are two points (-3,4),(1,7) What is the slope of the line between them?

valentina_108 [34]

Answer:

3/4

Step-by-step explanation:

3 0

3 years ago

When x increases from a to a + 2, y increases by a difference of 8.

nexus9112 [7]

Answer:

Step-by-step explanation: the answer is C y= 4x + 2 (usatestprep)

7 0

4 years ago

Read 2 more answers

In 2000 the total amount of gamma ray bursts was recorded at 6.4 million for a city. In 2005, the same survey was made and the t

ratelena [41]

Let's assume

It started in 2000

so, t=0 in 2000

$P_0=6.4million$

we can use formula

$P(t)=P_0 e^{rt}$

we can plug value

$P(t)=6.4 e^{rt}$

In 2005, the same survey was made and the total amount of gamma ray bursts was 7.3 million

so, at t=2005-2000=5

P(t)=7.3 million

we can plug value and then we can solve for r

$7.3=6.4 e^{r*5}$

$r=0.02632$

now, we can plug back

$P(t)=6.4 e^{0.02632t}$

now, we have

P(t)=1 billion =1000 million

so, we can set it and then we can solve for t

$1000=6.4 e^{0.02632t}$

$t=191.924$

approximately

$t=192$

Year is 2000+192

year is 2192................Answer

4 0

3 years ago