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2 years ago

Clarissa sells handmade bracelets for each. The graph below shows her profits after expenses.

Mathematics

2 answers:

LenKa [72]2 years ago

4 0

Answer:

where's the graph?

Step-by-step explanation:

kari74 [83]2 years ago

3 0

Answer:

Someone for real deleted my answer just because they got mad that I said theres no graph, I'm dead IMAO XD

Step-by-step explanation:

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The circumference of a circular pond in the park is 120 meters. Find the area.

Tpy6a [65]

Answer:

1) 38

2) 19

3) 1134

Step-by-step explanation:

7 0

2 years ago

What is the center of the hyperbola whose equation is (y+3)^2/81-(x-6)^2/89=1?

xxMikexx [17]

We have been given an equation of hyperbola $\frac{(y+3)^2}{81}-\frac{(x-6)^2}{89}=1$ . We are asked to find the center of hyperbola.

We know that standard equation of a vertical hyperbola is in form $\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$ , where point (h,k) represents center of hyperbola.

Upon comparing our given equation with standard vertical hyperbola, we can see that the value of h is 6.

To find the value of k, we need to rewrite our equation as:

$\frac{(y-(-3))^2}{81}-\frac{(x-6)^2}{89}=1$

Now we can see that value of k is $-3$ . Therefore, the vertex of given hyperbola will be at point $(6,-3)$ and option D is the correct choice.

6 0

2 years ago

Nicole recorded the balance of her loan, y, over a period of several months, x, in the table. Which scatter plot shows Х у 0 $3,

maxonik [38]

Answer:

it's A on PLATO

Step-by-step explanation:

5 0

2 years ago

Seed are often treated with fungicides to protect them in poor draining, wet environments. A small-scale trial, involving five t

Natalija [7]

Answer:

2.38%

Step-by-step explanation:

Given that:

The population size involving numbers of treated and untreated seeds N = 10

the number of seeds that are treated which are observed = 5

the number of seeds that sprouted 'n' = 4

Now, use the method of hypergeometric probability, we have the formula:

$p(y) = \dfrac{(^r_y)(^{N-r}_{n-y})}{(^N_n)}$

∴

$p(4) = \dfrac{(^5_4)(^{5}_{0})}{(^{10}_4)}$

$p(4) = \dfrac{\dfrac{5!}{4!(5-4)!}* \dfrac{5!}{0!(5-0)!} }{\dfrac{10!}{4!(10-4)!} }$

$p(4) = \dfrac{\dfrac{5!}{4!}* \dfrac{5!}{(5)!} }{\dfrac{10!}{4!(6)!} }$

$p(4) = \dfrac{5*1 }{210}$

p(4) = 0.0238

p(4) = 2.38%

3 0

3 years ago

The number of pollinated flowers as a function of time in days can be represented by the function. f(x)

ratelena [41]

The average increase in the number of flowers pollinated per day between days 4 and 10 is <u>39</u>, given that the number of pollinated flowers as a function of time in days can be represented by the function $f(x) = (3)^{\frac{x}{2} }$ .

In the question, we are asked for the average increase in the number of flowers pollinated per day between days 4 and 10, given that the number of pollinated flowers as a function of time in days can be represented by the function $f(x) = (3)^{\frac{x}{2} }$ .

To find the average increase in the number of flowers pollinated per day between days 4 and 10, we use the formula {f(10) - f(4)}/{10 - 4}.

First, we find the value of the function $f(x) = (3)^{\frac{x}{2} }$ , for f(10) and f(4).

$f(x) = (3)^{\frac{x}{2} }\\\Rightarrow f(10) = (3)^{\frac{10}{2} }\\\Rightarrow f(10) = 3^5 = 243$

$f(x) = (3)^{\frac{x}{2} }\\\Rightarrow f(4) = (3)^{\frac{4}{2} }\\\Rightarrow f(10) = 3^2 = 9$

Thus, the average increase

= {f(10) - f(4)}/{10 - 4},

= (243 - 9)/(10 - 4),

= 234/6

= 39.

Thus, the average increase in the number of flowers pollinated per day between days 4 and 10 is <u>39</u>, given that the number of pollinated flowers as a function of time in days can be represented by the function $f(x) = (3)^{\frac{x}{2} }$ .

Learn more about the average increase in a function at

brainly.com/question/7590517

#SPJ4

For complete question, refer to the attachment.

8 0

1 year ago