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Diano4ka-milaya [45]
2 years ago
14

Calvin Butterball is driving along the highway. He starts up a long, straight hill. 114 meters from the bottom of the hill Calvi

n's car runs out of gas. He doesn't put on the brakes, so the car keeps rolling for awhile, coasts to a stop, then start rolling backwards. He finds that his distances from the bottom of the hill 4 and 6 seconds after he runs out of gas are 198 and 234 meters, respectively. b.Write an equation expressing Calvin's distance from the bottom of the hill in terms of the number of seconds which have elapsed since he ran out of gas.
Mathematics
1 answer:
lbvjy [14]2 years ago
4 0
The answer is 6hours in 122 minutes do umknow
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ANSWER ALL 5 PARTS.
N76 [4]
A function f from a set A to a set B is defined as a relation that assings to each element  x in the set A exactly one element y in the set B. The set A is called the domain of the function while the set B is the range. So we have five statements and need to find some functions. Melissa decides to reserve a patch in her vegetable garden for growing bell peppers. If each side of the tomato patch is x feet, then we have a square patch as shown in the Figure below.

1.a) Write the function Wa(x) representing the width of the bell pepper patch.

We know that she wants its width to be half the width of the tomato patch. Let x be the width of the tomato patch, then the function that matches this statement is:

\boxed{Wa(x)=\frac{x}{2}}

1.b) Write the function La(x) representing the length of the bell pepper patch.

In this case Melissa wants <span>its length to exceed the length of the tomato patch by 2 feet. To do this we enlarge the length of the tomato patch 2 feet. Therefore the function is the following:

</span>\boxed{La(x)=x+2}
<span>
2. Ar</span>ea of the bell pepper patch in terms of x.

Given that the bell pepper patch is a rectangle, then t<span>he area of a rectangle is the product of the length and width. So:

</span>A=(\frac{x}{2})(x+2) \\ \\ \therefore \boxed{A=\frac{x(x+2)}{2}}
<span>
3. C</span><span>ombined area of the tomato patch and the bell pepper patch.

This function is the sum of both the area of the tomato patch and the bell pepper patch. So:

</span>Aar(x)=x^{2}+\frac{x(x+2)}{2} \rightarrow Aar(x)=x^{2}+ \frac{x^{2}}{2}+x \rightarrow Aar(x)=\frac{3x^{2}}{2}+x \\ \\ \therefore \boxed{Aar(x)=\frac{x(3x+2)}{2}}
<span>
4. W</span>rite the function Aa(x) for the remaining planting area in the garden.

The remaining planting area in the garden are the rectangles in red. So we need to subtract the width of the bell pepper patch from the width of the tomato patch and multiply it by 2. In mathematical language this is given by:<span>

</span>Aa(x)=2(x-\frac{x}{2}) \rightarrow Aa(x)=x

5. 
Find the area of the remaining space in the garden after planting tomatoes and bell peppers.

Given that <span>Melissa wants the area of the bell pepper patch to be 31.5 square feet, then it is true that:

</span>31.5=\frac{x(x+2)}{2} \rightarrow x^{2}+2x-64=0 \\ \\ solving \ for \ x: \\ x=7.06
<span>
Therefore the area of the remaining space is:

</span>\boxed{Aa(7.06)=7.06ft^{2}}

6 0
2 years ago
Which statement is true about the graphed function
natka813 [3]

From -∞ to -4 the blue line is above the X axis which means it is >0

The blue line is negative between -4 and -3


This would make the correct answer: F(x) > 0 over the interval (-∞,-4)

8 0
2 years ago
Read 2 more answers
The temperature in degrees Kelvin is 273 degrees less than the temperature in degrees Celsius. Use the formula LaTeX: C=\frac{5}
katrin2010 [14]

Answer:

°F = 9/5 (°K - 273) + 32

Step-by-step explanation:

°C = °K - 273

°F = 9/5 °C + 32

Put the 1st formula in the 2nd

°F = 9/5 (°K - 273) + 32

5 0
2 years ago
Read 2 more answers
A basketball court is a rectangle that is 28\,\text{m}28m28, start text, m, end text long and 15\,\text{m}15m15, start text, m,
vladimir2022 [97]

Answer:

48

Step-by-step explanation:

7 0
2 years ago
HELP ASAP In two or more complete sentences, Explain how you would find the equation of a parabola, given the coordinate of the
kvasek [131]
In the general case in Cartesian coordinates, you would use the definition of a parabola as the locus of points equidistant from the focus and directrix. The equation would equate the square of the distance from a general point (x, y) to the focus with the square of the distance from that point to the directrix line.

Suppose the focus is located at (h, k) and the equation of the directrix is ax+by+c=0. The expression for the square of the distance from (x, y) to the point (h, k) is ...
  (d₁)² = (x-h)²+(y-k)²
The expression for the square of the distance from (x, y) to the directrix line is
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Equating these expressions gives the equation of the parabola.
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When the directrix is parallel with one of the axes, one of the coefficents "a" or "b" is zero and the equation becomes much simpler. Often, it would be easier to make use of the formula (for a directrix parallel to the x-axis):
  y = 1/(4p)*(x -h)² +k
where the (h, k) here is the vertex, the point halfway between the focus and directrix, and "p" is the (signed) distance from the focus to the vertex. (p is positive when the focus is above or to the right of the vertex.)
4 0
2 years ago
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