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Karo-lina-s [1.5K]
4 years ago
7

A function is represented by the graph.

Mathematics
2 answers:
DiKsa [7]4 years ago
7 0

Answer:

'Equal to' is the answer.

Step-by-step explanation:

The equation of a straight line is given by y=mx+c, where m= slope and c= y-intercept.

In the given equation y=3x+2, the y-intercept is 2.

Now, from the graph we get two pair of points (0,2) and (-1,0). Using these points, we will find the slope.

i.e. m = \frac{y_{2}-y_{1}}{x_{2} -x_{1}}

i.e. m= \frac{0-2}{-1-0}

i.e. m= 2.

Now, using this slope and the point (-1,0) in the given equation, we will find the equation of the graph.

i.e. (y-y_{1}) = m*(x-x_{1})

i.e y-0 = 2*(x-(-1))

i.e. y= 2*(x+1)

i.e. y= 2x+2

Therefore, y=2x+2 is the equation of the graph and comparing with the general form at the start, the y-intercept is 2.

Hence, both equations have equal y-intercept.



WITCHER [35]4 years ago
3 0
<span>equal to >>>>>>>>>>>>>>>>>>>>>>>>>
</span>
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gladu [14]

Answer:

Graph (a)

Step-by-step explanation:

When we take x = 0 and replace in the given formula we get:

y = 0.5 (0.5)ˣ = 0.5 (0.5)⁰ = 0.5 · 1 = 0.5  and this corresponds to graph (a).

God with you!!!

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3 years ago
In the given figure, if a:b=10:13, then /_xoy equals.​
soldier1979 [14.2K]

Answer:

2x + 7 = 3x - 7

3x - 2x = 7 + 7

x = 14

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2(14) + 12y + 8 = 180

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Ivanshal [37]

Answer:

t = 17.5 f = 12

Step-by-step explanation:

If <em>MNOP </em>is a square, you know that each angle is going to be 90°. To solve, let's first set each equation equal to 90°:

90 = 4t + 20

and

90 = 7f + 6

Then solve both equations:

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3 years ago
If an object is dropped from a height of 55 feet, the function d = -16^2 + 55 gives the height of the object after t seconds. Gr
motikmotik

Answer:

Approximately  1.9 seconds   (correct to nearest tenth)

Step-by-step explanation:

Looks like the function is  d = -16t^2 + 55    ( you left out the t)

The answer is the value of t when d = 0 so we have the equation:-

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Answer:

$6960

Step-by-step explanation:

p is not the principle. It is the principal.

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