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timama [110]
4 years ago
9

The dimensions of a big dog house are 2 1/2 the dimensions of a small dog house. If the width of the small doghouse is 2 feet, w

hat is the width of the big dog house?
Mathematics
1 answer:
Alchen [17]4 years ago
6 0
   The width of the big dog house is  2.5 times the small dog house. 
   multiply 2 by 2.5 to get your answer. hope this helps :)
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Find the value for x. Round your answer to the nearest hundredth. Area of the triangle = 8 ft^2​
zmey [24]

Answer:

x = 3.51

Step-by-step explanation:

Since, formula to determine the area f a triangle is,

Area = \frac{1}{2}(\text{Base})(\text{Height})

     8 = \frac{1}{2}(x + 1)(3x - 7)

16 = x(3x - 7) + 1(3x - 7)

16 = 3x² - 7x + 3x - 7

16 = 3x² - 4x - 7

0 = 3x² - 4x - 23

3x²- 4x - 23 = 0

By quadratic formula,

x = \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}

x = \frac{4\pm \sqrt{(-4)^{2}-4(3)(-23)}}{2(3)}

x = \frac{4\pm \sqrt{292}}{6}

x = \frac{4\pm 17.09}{6}

x = 3.51, -2.18

But the length of sides can't be negative.

Therefore, x = 3.51 will be the answer.

8 0
3 years ago
Write an equation of a line that passes through the point (1, 2) and is perpendicular to the line y=-1/4x+2​
Dmitriy789 [7]

Answer:

y = 4x - 2 would be the equation

7 0
3 years ago
Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following p
Lisa [10]

Answer:

0.0064

0.00032

Step-by-step explanation:

Given the details:

P(X > 3), n = 5, p = 0.2

The binomial distribution is related using the formula:

P(x = x) = nCx * p^x * q^(n-x)

q = 1 - p = 1 - 0.2 = 0.8

P(X > 3) = p(x = 4) + p(x = 5)

P(x = 4) = 5C4 * 0.2^4 * 0.8^1 = 5 * 0.2^4 * 0.8^1 = 0.0064

P(x = 5) = 5C5 * 0.2^5 * 0.8^0 = 1 * 0.2^5 * 0.8^0 = 0.00032

3 0
3 years ago
Write a function with the following characteristics: 1.A vertical asymptote at x = 3 A horizontal asymptote at y = 2 An x-interc
elena-14-01-66 [18.8K]

1. The vertical asymptote requires the denominator have a zero at that location. The x-intercept requires the numerator have a zero at that location. The horizontal asymptote amounts to a multiplier of the function:

... y = 2(x +5)/(x -3)

2. The vertical asymptote requires the denominator have a zero at that location. The oblique asymptote is an add-on

... y = 1/(x +1) +(x +2)

... y = (x² +3x +3)/(x +1)

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3 years ago
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