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3 years ago

14

Identify the domain and range for (1,4) (2,5) (0,6) (1,7) (2,8)

1 answer:

melamori03 [73]3 years ago

7 0

The domain are the x values: 0,1,2
The range are the y values: 4,5,6,7,8

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Questions attached as screenshot below:Please help me I need good explanations before final testI pay attention

Nikitich [7]

The acceleration of the particle is given by the formula mentioned below:

$a=\frac{d^2s}{dt^2}$

Differentiate the position vector with respect to t.

$\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}$

Differentiate both sides of the obtained equation with respect to t.

$\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}$

Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.

$\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}$

The initial position is obtained at t=0. Substitute t=0 in the given position function.

$\begin{gathered} s(0)=-23\times0+65 \\ =65 \end{gathered}$

8 0

1 year ago

A boy is building a pyramid out of building blocks. He puts 20 blocks in the first row, then he puts 19 blocks in the row above,

rjkz [21]

Answer:

<h3>The answer is 206 I believe</h3>

8 0

4 years ago

If a=under root s(s-a)(s-b)(s-c) then the value of a, when a =3 b =4c = 5 and s =a+b+c by 2

kramer

Answer:

6

Step-by-step explanation:

Here it's given that ,

$\sf\red{\longrightarrow} A=\sqrt{s(s-a)(s-b)(s-c)}$

Also ,

$\sf\red{\longrightarrow} s =\dfrac{a+b+c}{2}$

And we need to find out the value of A , when

a = 3
b = 4
c = 5

So , on substituting the respective values to find s we have ,

$\sf\red{\longrightarrow} s =\dfrac{3+4+5}{2}=\dfrac{12}{2}=\bf{6}$

Now let's find out A as ,

$\sf\red{\longrightarrow} A = \sqrt{6(6-3)(6-4)(6-5)} \\$

$\sf\red{\longrightarrow}A =\sqrt{6 * 3 * 2 * 1}\\$

$\sf\red{\longrightarrow} A = \sqrt{3^2 * 2^2 *1^2}\\$

$\sf\red{\longrightarrow}A = 3*2\\$

$\sf\red{\longrightarrow}\boxed{\qquad \blue{\bf A = 6 \qquad}}$

7 0

2 years ago

What is the value of minus 6 x cubed minus y squared minus 3 x y if if x = -2 and y = 4

Oksi-84 [34.3K]

Well, "minus 6 x cubed minus y squared minus 3 x y" translates to:

$-6x^{2} -y^{2} -3xy$

Then, if we insert the values for x and y, we get:

$-6(-2)^{2} -(4)^{2} -3(-2)(4)$

When we distribute and multiply:

$144-16-24$

And once we combine like terms:

104<em> is the answer</em>

8 0

3 years ago

Read 2 more answers

. Two stacks of flashcards each contain a card with one of the digits 0-9 written on the card. What is the probability of drawin

gayaneshka [121]

1/5

Step-by-step explanation:

(2,6) = 4/20 = 1/5

answer is 1/5

7 0

2 years ago