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ch4aika [34]
3 years ago
8

Kim scored 33 points at her basketball game game with a combination of 2-point shots and 3-point shots. If she made a total of 1

5 baskets, how many of each kind oh shot did she make?
Mathematics
2 answers:
3241004551 [841]3 years ago
6 0

Answer:

3 3-point shots

12 2-point shots

Step-by-step explanation:

x = 2-point shots

y = 3-point shots

x + y = 15

2x + 3y = 33

x = 15-y

30-2y + 3y = 33

30 + y = 33

y = 3

x = 12

Rudik [331]3 years ago
3 0

Answer:

(12, 3)

Step-by-step explanation:

write the equation:

if x = number of 2 point shots

and y = number of 3 point shots

2x + 3y = 33

write another equation with the total baskets

x + y = 15

plug them both into desmos and find the solution

(12, 3)

she made 12 2-point shots and 3 3-point shots

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Answer:

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Step-by-step explanation:

divide 6 by 3 and multiply that answer by 10

8 0
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Step-by-step explanation:

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4 0
3 years ago
How is the theoretical probability of an event​ computed?
Sedaia [141]

The theoretical probability of an event​ computed as

Theoretical Probability = \frac{ Number of desired outcomes}{ Number of possible outcomes.} This is further explained below.

<h3>What is theoretical probability?</h3>

Generally, The theoretical probability formula is the ratio of the number of successful outcomes to the total number of possible possibilities. We may write this formula out as follows:

In conclusion, the theoretical probability is mathematically given as

Theoretical Probability = \frac{ Number of desired outcomes}{ Number of possible outcomes.}

Read more about theoretical probability

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6 0
2 years ago
Any help with an explanation would be appreciated!
Lilit [14]

Problem 1

We'll use the product rule to say

h(x) = f(x)*g(x)

h ' (x) = f ' (x)*g(x) + f(x)*g ' (x)

Then plug in x = 2 and use the table to fill in the rest

h ' (x) = f ' (x)*g(x) + f(x)*g ' (x)

h ' (2) = f ' (2)*g(2) + f(2)*g ' (2)

h ' (2) = 2*3 + 2*4

h ' (2) = 6 + 8

h ' (2) = 14

<h3>Answer: 14</h3>

============================================================

Problem 2

Now we'll use the quotient rule

h(x) = \frac{f(x)}{g(x)}\\\\h'(x) = \frac{f'(x)*g(x)-f(x)*g'(x)}{(g(x))^2}\\\\h'(2) = \frac{f'(2)*g(2)-f(2)*g'(2)}{(g(2))^2}\\\\h'(2) = \frac{2*3-2*4}{(3)^2}\\\\h'(2) = \frac{6-8}{9}\\\\h'(2) = -\frac{2}{9}\\\\

<h3>Answer:  -2/9</h3>

============================================================

Problem 3

Use the chain rule

h(x) = f(g(x))\\\\h'(x) = f'(g(x))*g'(x)\\\\h'(2) = f'(g(2))*g'(2)\\\\h'(2) = f'(3)*g'(2)\\\\h'(2) = 3*4\\\\h'(2) = 12\\\\

<h3>Answer:  12</h3>
5 0
3 years ago
F(x)=2x^2-x-10 part a: what are the X intercepts of the graph of f(x)? Show your work
Genrish500 [490]
A. x intercepts are where the graph hits the x axis or where f(x)=0
0=2x^2-x-10
solve
hmm
we can use the ac method
multiply 2 and -10 to get -20
what 2 numbers muliply to get -20 and add to get -1 (the coefint of the x term)
-5 and 4
split the midd into that
0=2x^2+4x-5x-10
group
0=(2x^2+4x)+(-5x-10)
factor
0=2x(x+2)-5(x+2)
undistribute
0=(x+2)(2x-5)
set each to 0
0=x+2
0=-2

0=2x-5
5=2x
5/2=x

x intercepts are at x=-2 and 5/2 or the points (-2,0) and (5/2,0)


B. ok, so for f(x)=ax^2+bx+c
if a>0, then the parabola opens up and the vertex is a minimum
if a<0 then the parabola opens down and the vertex is a max

f(x)=2x^2-x-10
2>0
opoens up
vertex is minimum

ok, the vertex

the x value of  the vertex in f(x)=ax^2+bx+c= is -b/(2a)
the y value of the vertex is f(-b/(2a)) so
given
f(x)=2x^2-x-10
a=2
b=-1
-b/2a=-(-1)/(2*2)=1/4
f(1/4)=2(1/4)^2-(1/4)-10
f(1/4)=2(1/16)-1/4-10
f(1/4)=1/8-1/4-10
f(1/4)=1/8-2/8-80/8
f(1/4)=-81/8
so the vertex is (1/4,-81/8) or (0.25,-10.125)


C. graph the x intercepts and the vertex
the vertex is min and the graph goes through the x intercepts
4 0
3 years ago
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