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valentina_108 [34]
3 years ago
13

The scale model of a rectangular garden is 1.5 ft by 4 ft. The scale model is enlarged by a scale factor of 7 to create the actu

al garden. What is the area of the actual garden?
6 ft2

42 ft2

252 ft2

294 ft2
Mathematics
2 answers:
LenaWriter [7]3 years ago
6 0

It is given that the scale model of a rectangular garden is 1.5 ft by 4 ft. The scale model is enlarged by a scale factor of 7 to create the actual garden.

Therefore, we can see clearly that the width of the scale model is 1.5 feet. Hence, the width of the actual garden which has been enlarged by a scale factor of 7 will be 7 times the width of the scale model.

Thus the width of the actual garden will be: 1.5\times 7=10.5 feet

In a similar fashion the length of the actual garden will be 4\times 7=28 feet

Thus, the area of the actual garden will be:

Area=Length\times width=10.5\times 28=294 ft^2

As we can see, out of the given options, the last option is the correct one.

nika2105 [10]3 years ago
6 0

Answer:

294 ft2

Step-by-step explanation:

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Which statement describes the inverse of m(x) = x2 – 17x?
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Answer:

The correct option is;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

Step-by-step explanation:

The given information is that m(x) = x² - 17·x

The above equation can be written in the form;

y = x² - 17·x

Therefore;

0 = x² - 17·x - y

From the general solution of a quadratic equation, 0 = a·x² + b·x + c we have;

x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}

By comparison to the equation,0 = x² - 17·x - y, we have;

a = 1, b = -17, and c = -y

Substituting the values of a, b and c into the formula for the general solution of a quadratic equation, we have;

x = \dfrac{-(-17)\pm \sqrt{(-17)^{2}-4\times (1) \times (-y)}}{2\times (1)} = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}

Which can be simplified as follows;

x =  \dfrac{17\pm \sqrt{289+4\cdot y}}{2}= \dfrac{17}{2} \pm \dfrac{1}{2}  \times \sqrt{289+4\cdot y}} = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +\dfrac{4\cdot y}{4} }}

And further simplified as follows;

x = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +y }} = \dfrac{17}{2} \pm \sqrt{y + \dfrac{289}{4} }}

Interchanging x and y in the function of the inverse, m⁻¹(x), we have;

m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

We note that the maximum or minimum point of the function, m(x) = x² - 17·x found by differentiating the function and equating the result to zero, gives;

m'(x) = 2·x - 17 = 0

x = 17/2

Similarly, the second derivative is taken to determine if the given point is a maximum or minimum point as follows;

m''(x) = 2 > 0, therefore, the point is a minimum point on the graph

Therefore, as x increases past the minimum point of 17/2, m⁻¹(x) increases to give;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }} to increase m⁻¹(x) above the minimum.

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Hey! I would appreciate some help with this
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Answer:

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Step-by-step explanation:

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i see u dont want it solved but ill solve it anyways

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