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elena55 [62]
3 years ago
8

Which of the following equations will produce the graph shown below?

Mathematics
2 answers:
melisa1 [442]3 years ago
5 0
The graph is a circle, centered at the origin, with radius=4.

We know that we can write the equation of a circle with radius r and center (a,b) as :

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

Thus, substituting (a, b) with (0, 0) and r with 4, we have:

                      x^2+y^2=16.

The solutions of this equation are all the points forming the circle shown in the picture. The solutions of this equation are still the same even if we multiply both sides of the equation by 2, because rewriting the equation as:

x^2+y^2=16\\\\x^2+y^2-16=0\\\\2(x^2+y^2-16)=0, 

we would still have the same roots.

Thus, the equation of the circle can be written as :

                                    2x^2+2y^2=32.


Answer: B

Gnesinka [82]3 years ago
3 0

Answer:

B

Step-by-step explanation:

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At the corner market it costs $4.32 for a dozen apples. How much would it cost if I bought five apples?
matrenka [14]

Answer:

so do $4.32 divided that by 12 and you will get a new number and then you take ur new number and times that by 5.

7 0
3 years ago
Consider the parabola r​(t)equalsleft angle at squared plus 1 comma t right angle​, for minusinfinityless thantless thaninfinity
kodGreya [7K]

Given:-   r(t)=< at^2+1,t>  ; -\infty < t< \infty , where a is any positive real number.

Consider the helix parabolic equation :  

                                              r(t)=< at^2+1,t>

now, take the derivatives we get;

                                            r{}'(t)=

As, we know that two vectors are orthogonal if their dot product is zero.

Here,  r(t) and r{}'(t)  are orthogonal i.e,   r\cdot r{}'=0

Therefore, we have ,

                                  < at^2+1,t>\cdot < 2at,1>=0

< at^2+1,t>\cdot < 2at,1>=

                                              =2a^2t^3+2at+t

2a^2t^3+2at+t=0

take t common in above equation we get,

t\cdot \left (2a^2t^2+2a+1\right )=0

⇒t=0 or 2a^2t^2+2a+1=0

To find the solution for t;

take 2a^2t^2+2a+1=0

The numberD = b^2 -4ac determined from the coefficients of the equation ax^2 + bx + c = 0.

The determinant D=0-4(2a^2)(2a+1)=-8a^2\cdot(2a+1)

Since, for any positive value of a determinant is negative.

Therefore, there is no solution.

The only solution, we have t=0.

Hence, we have only one points on the parabola  r(t)=< at^2+1,t> i.e <1,0>




                                               




6 0
3 years ago
M(-5, 2) and N(5, 2) are the endpoints of the segment MN on the coordinate plane. What is the length of ?
VikaD [51]
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6 0
3 years ago
Read 2 more answers
A phone company offers two monthly plans. Plan A costs $17 plus an additional $0.09 for each minute of calls. Plan B costs $22 p
Varvara68 [4.7K]

Answer: $53

Step-by-step explanation:

17+ .09m = 9 + .11m

8 = .02m

m = 8/.02= 400 minutes for the two plans to cost the same

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8 0
3 years ago
A college job placement office collected data about students’ GPAs and the salaries they earned in their first jobs after gradua
frez [133]

Answer:

X is the GPA

Y is the Salary

Standard deviation of X is 0.4

Standard deviation of Y is 8500

E(X)=2.9

E(Y)=47200

We are given that The correlation between the two variables was r = 0.72

a)y = a+bx

b = \frac{\sum(x_i-\bar{x})(y_i-\bar{y})}{\sum(x_i-\bar{x})^2} = \frac{r \times \sqrt{var(X) \times Var(Y)}}{Var(X)} =  \frac{0.72 \times \sqrt{0.4^2 \times 8500^2}}{0.4^2} = 15300

a=y-bx = 47200-(15300 \times 29) = 2830

So, slope =  15300

Intercept =  2830

So, equation : y = 2830+15300x

b) Your brother just graduated from that college with a GPA of 3.30. He tells you that based on this model the residual for his pay is -$1880. What salary is he earning?

y = 2830+15300 \times 3.3 = 53320

Observed salary = Residual + predicted = -1860+53320 = 51440

c)) What proportion of the variation in salaries is explained by variation in GPA?

The proportion of the variation in salaries is explained by variation in GPA = r^2 = (0.72)^2 =0.5184

8 0
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