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3 years ago

This is due at midnight someone helppppppppp

Mathematics

2 answers:

Dmitry_Shevchenko [17]3 years ago

7 0

Answer:

x,y= 1,3

Step-by-step explanation:

x = 1

y = 3

x,y = 1,3

Alik [6]3 years ago

3 0

X = 1
and = 3
(x,and) = (1,3)

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Answer:

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3 years ago

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qaws [65]

Answer:

Point C is located on the x-axis.

Step-by-step explanation:

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3 years ago

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Drag each tile to the correct box.

Natasha_Volkova [10]

Answer:

1) Function h

interval [3, 5]

rate of change 6

2) Function f

interval [3, 6]

rate of change 8.33

3) Function g

interval [2, 3]

rate of change 9.6

Step-by-step explanation:

we know that

To find the average rate of change, we divide the change in the output value by the change in the input value

the average rate of change is equal to

$\frac{f(b)-f(a)}{b-a}$

step 1

Find the average rate of change of function h(x) over interval [3,5]

Looking at the third picture (table)

$f(a)=h(3)=4$

$f(b)=h(5)=16$

$a=3$

$b=5$

Substitute

$\frac{16-4}{5-3}=6$

step 2

Find the average rate of change of function f(x) over interval [3,6]

Looking at the graph

$f(a)=f(3)=10$

$f(b)=f(6)=35$

$a=3$

$b=6$

Substitute

$\frac{35-10}{6-3}=8.33$

step 3

Find the average rate of change of function g(x) over interval [2,3]

we have

$g(x)=\frac{1}{5}(4)^x$

$f(a)=g(2)=\frac{1}{5}(4)^2=\frac{16}{5}$

$f(b)=g(3)=\frac{1}{5}(4)^3=\frac{64}{5}$

$a=2$

$b=3$

Substitute

$\frac{\frac{64}{5}-\frac{16}{5}}{3-2}=9.6$

therefore

In order from least to greatest according to their average rates of change over those intervals

1) Function h

interval [3, 5]

rate of change 6

2) Function f

interval [3, 6]

rate of change 8.33

3) Function g

interval [2, 3]

rate of change 9.6

7 0

3 years ago

Find a compact form for generating functions of the sequence 1, 8,27,... , k^3

pantera1 [17]

This sequence has generating function

$F(x)=\displaystyle\sum_{k\ge0}k^3x^k$

(if we include $k=0$ for a moment)

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k$

Take the derivative to get

$\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k$

$\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k$

Take the derivative again:

$\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k$

$\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k$

Take the derivative one more time:

$\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k$

$\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k$

so we have

$\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}$

5 0

3 years ago

A bag contains 5 black marbles, 7 blue marbles, 3 green marbles, 4 red marbles, and 1 white marble. A marble is selected and the

schepotkina [342]

Answer:

2/25

Step-by-step explanation:

It will be 20/400 + 12/400

=2/25

5 0

3 years ago