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4 years ago

9

Consider the graph.

2 answers:

Arlecino [84]4 years ago

5 0

Answer:

C, f(x) = 2x + 6

Step-by-step explanation:

First, we need to plug in the values of the x coordinates and see if it matches with the y coordinate to determine if it is on the same line. Startin with 2x + 8, we have the point (1, 8) on the graph. Plugging in 1 gets you 10 for the y. This is wrong since 8 is the y coordinate. Moving on, we have 6.4(1.25)^x for the same point. Plugging in 1, we have 6.4 * 1.25 = 8, which is true. Moving on to the second point, (2, 10), we have 1.25 squared times 6.4. This is thus wrong. So, moving on to 2x + 6, we have the point (1, 8), and plugging in 1 for x, we have 8 as y. Since this satisfies the equation we move on to the next point, (2,10). Plugging in x, we have 2 * 2 + 6 = 10, which is also true. Moving on to our third point (3 , 12), we plug in 3 for x. We then get 3 * 2 + 6 = 12, which is correct. This, is our answer then.

kodGreya [7K]4 years ago

4 0

Answer:

f(x) = 2x + 6

Step-by-step explanation:

Well this is actually pretty simple we can easily find the slope from see how far away each points are from each other which is 2x.

So now we have to find b or the y intercept which is easily found by using the slope and counting backwards. When doing that we can tell that b is 6.

So the answer is y = 2x + 6.

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3 years ago

Two buildings on opposites sides of a highway are 3x^3- x^2 + 7x +100 feet apart. One building is 2x^2 + 7x feet from the highwa

Furkat [3]

Given:

Distance between two buildings = $3x^3- x^2 + 7x +100$ feet apart.

Distance between highway and one building = $2x^2 + 7x$ feet.

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To find:

The standard form of the polynomial representing the width of the highway between the two building.

Solution:

We know that,

Width of the highway = Distance between two buildings - Distance of both buildings from highway.

Using the above formula, we get the polynomial for width (W) of the highway.

$W=3x^3- x^2 + 7x +100-(2x^2 + 7x)-(x^3 + 2x^2 - 18)$

$W=3x^3- x^2 + 7x +100-2x^2-7x-x^3 -2x^2+18$

Combining like terms, we get

$W=(3x^3-x^3)+(- x^2 -2x^2-2x^2)+ (7x -7x)+(100 +18)$

$W=2x^3-5x^2+0+118$

$W=2x^3-5x^2+118$

Therefore, the width point highway is $2x^3-5x^2+118$ .

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3 years ago

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Likurg_2 [28]

Answer:

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Step-by-step explanation:

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ruslelena [56]

Answer:

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Step-by-step explanation:

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andreev551 [17]

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Step-by-step explanation:

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4 years ago

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