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3 years ago

Calculate the shaded areas for c and d

Mathematics

1 answer:

KiRa [710]3 years ago

5 0

C)
Area of full triangle
1/2(8)(6)=24cm^2
Area of non shaded triangle
1/2(4)(3)=6cm^2
Area of full minus area of small triangle
24-6=18cm^2

D)
Area of Big rectangle
(4)(12)=48
Area of non shaded triangle
1/2(9)(4)=18
Area of Big rectangle minus non shaded triangle
48-18=30cm^2

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HELP TIMER Write the equation of a hyperbola centered at the origin with x-intercept +/- 4 and foci of +/-2(squareroot 5)

nikitadnepr [17]

Answer:

$\frac{x2}{a} - \frac{y2}{b2} = 1$

Step-by-step explanation:

A hyperbola is the locus of a point such that its distance from a point to two points (known as foci) is a positive constant.

The standard equation of a hyperbola centered at the origin with transverse on the x axis is given as:

$\frac{X2}{16} - \frac{b}{4} = 1$

The coordinates of the foci is at (±c, 0), where c² = a² + b²

Given that a hyperbola centered at the origin with x-intercepts +/- 4 and foci of +/-2√5. Since the x intercept is ±4, this means that at y = 0, x = 4. Substituting in the standard equation:

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a. = 4

The foci c is at +/-2√5, using c² = a² + b²:

B = 2

Substituting the value of a and b to get the equation of the hyperbola:

$\frac{x2}{a2} - \frac{y2}{b2} = 1$

$\frac{x2}{16} - \frac{b2}{4} = 1$

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3 years ago

Simplify in Exponential form <br> First correct answer gets BRAINLIEST

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Answer:

Steps to simplifying fractions

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4 years ago

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Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10

zvonat [6]

The approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

$FV=P\left(1-\dfrac{r}{100}\right)^n$

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Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is $n_1$ . Thus, by the above formula for the first car,

$0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}$

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Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is $n_2$ . Thus, by the above formula for the second car,

$0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}$

Take log both the sides as,

$\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14$

The difference in the ages of the two cars is,

$d=4.85-3.14\\d=1.71\rm years$

Thus, the approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.

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