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finlep [7]
3 years ago
12

Jane threw a ball into the air. The graph below shows the relationship between the number of seconds that the ball was in the ai

r and the height of the ball.
What is the x-intercept for the given function? What does this represent in the situation?


look at the graph ​

Mathematics
2 answers:
nadya68 [22]3 years ago
7 0

Answer:

the x intercept would be (2.5,0) and it would represent the moment the ball hit the ground or was no longer airborne

Nataly [62]3 years ago
7 0

Answer:

The x intercept is where the x appears on y axis (crosses line) As the line appears vertical and increases up x becomes time greater than this.

Upon the axis intercept we can see the ball ; it is stationary and thrown up as time moves across.

Step-by-step explanation:

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Answer:

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If we want to fit a linear regression through these seven observations, the slope would be positive as linear regression line passes through the average.

Looking at the relationship, slope must be significantly different from zero as there is a relationship between two variables, where as slope significantly zero implies no relationship. However, we have only seven points so drawing a conclusion is difficult as the nature of the relationship is not clearly visualized by this graph.

8 0
4 years ago
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Riley is 3 years older than his sister. What is the expression
Svetach [21]
X+3=age i believe lol
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3 years ago
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What’s the answer of this math problem
Klio2033 [76]
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4 years ago
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f of x equals the quantity x minus one times the quantity x plus two times the quantity x plus four all divided by the quantity
Alik [6]
f(x)=\frac{(x-1)(x+2)(x+4)}{(x+1)(x-2)(x-4)}

The denominator of a fraction can't be equal to 0.
(x+1)(x-2)(x-4) \not= 0 \\
x+1 \not=0 \ \land \ x-2 \not= 0 \ \land \ x-4 \not= 0 \\
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7 0
4 years ago
the weight of an object on mars varies directly as the weight of the object on earth a 90 pound object on earth weighs 34 pounds
Mariulka [41]

We have been given that the weight of an object on the mars varies directly as the weight of the object on earth.  

Let y be weight of an object on Mars and x be weight of an object on Earth.

\\
y=kx

We have been given that the weight of an object on Mars is 34 pounds and weight of the same object on Earth is 90 pounds.

Putting the given values in equation we get

\\
34=k(90)\\
\\
k=\frac{34}{90}=\frac{17}{45}

Therefore, our equation now becomes

\\
y= \frac{17}{45}x

Now we have been given that the weight of an object on Earth is 135 pounds and we have to find the weight of the same object on Mars,

We have been given that \\
x=135

Putting the given values in our equation we get

\\
y= \frac{17}{45}(135)=\frac{17\cdot 135}{45}=17\cdot 3 = 51

Therefore weight of the object on Mars will be 51 pounds.

                                                                             

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