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Ainat [17]
2 years ago
9

Identify the equation of the circle that has its center at (10, 24) and passes through the origin.

Mathematics
1 answer:
marishachu [46]2 years ago
5 0

Answer:

A )  The Equation of the circle (x−10)2+(y−24)2=676

Step-by-step explanation:

step1:-

The equation of the circle whose center is (a,b) and radius r is

(x-a)^{2} +(y-b)^{2} =r^{2}

(x-10)^{2}+(y-24)^{2}  =676

in this circle equation centre is (g,f) = (10,24)

and formula of radius of a circle is

r = \sqrt{g^{2}+f^{2} -c }

Step2:-

The Equation of the circle (x−10)^2+(y−24)^2=676

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In a small town, 50% of single family homes have a front porch. 48 single family homes are randomly selected. Let X represent th
Reil [10]

Answer:

N(24, 3.46)

Step-by-step explanation:

We are given the following in the question:

Percentage of family homes having front poach = 50%

p = 50\% = 0.50

Sample size, n = 48

Normal approximation to the given distribution:

\mu = np = 48(0.50) = 24

\sigma = \sqrt{np(1-p)} = \sqrt{48(0.50)(1-0.50)} = 3.46

Thus, the distribution of single family homes is best approximated by the normal distribution N(24, 3.46) where mean is 24 and standard deviation is 3.46

6 0
2 years ago
If the end behavior is increasing to the left, what might be true about the function? Select all that apply.
slava [35]

The degree is even and the leading coefficient is positive

The degree is odd and the leading coefficient is negative

Step by step: Trust me

6 0
2 years ago
Assemble the proof by dragging tiles to the statements and reasons columns.<br><br> PLEASE HELP
nataly862011 [7]

m<ABD = 52                               Given

BC bisects <ABD                        Given

m<ABC =  m<CBD                      Definition of angle bisector.

m<ABC +m<CBD = m<ABD        Angle addition postulate.

m<ABC +m<CBD = 52                Substitution

m<ABC + m<ABC= 52                Substitution

2m<ABC = 52                             Combining like angles

m<ABC = 26                                Division property of equality.  

5 0
2 years ago
Read 2 more answers
the area of a rectangle banquet hall is 7400 square feet. the length of one side of the hall is 82 feet. explain how you can use
Vadim26 [7]
We are given that the hall has a rectangular shape.
Area of rectangle can be calculated using the following rule:
Area of rectangle = length * width

We are also given that:
Area = 7400 square feet
length = 82 feet

Substitute with the givens in the above equation to get the width as follows:
Area of rectangle = length * width
7400 = 82 * width
width = 7400 / 82
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5 0
2 years ago
What are the zeros of the quadratic function f(x) = 6x^2 + 12x – 7?
strojnjashka [21]

Answer:

x1=\frac{-2+\sqrt{26/3}}{2}

x2=\frac{-2-\sqrt{26/3} }{2}

Step-by-step explanation:

To find the zeros of the quadratic function f(x)=6x^2 + 12x – 7 we need to factorize the polynomial.

To do so, we need to use the quadratic formula, which states that the solution to any equation of the form ax^2 + bx + c = 0 is:

x=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

So, the first thing we're going to do is divide the whole function by 6:

6x^2 + 12x – 7 = 0 -> x^2 + 2x - 7/6

This step is optional, but it makes things quite easier.

Then we using the quadratic formula, where:

a=1, b= 2, c = -7/6.

Then:

x=\frac{-2±\sqrt{2^{2}-4(1)(-7/6)}}{2}

x=\frac{-2±\sqrt{4 +14/3}}{2}

x=\frac{-2±\sqrt{26/3}}{2}

So the zeros are:

x1=\frac{-2+\sqrt{26/3}}{2}

x2=\frac{-2-\sqrt{26/3}}{2}

4 0
2 years ago
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