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2 years ago

What is the avarage rate from x=1 to x=3

2 answers:

6 0

It depends on the line or parabola that you are talking about, more information would be needed to answer this correctly.

sineoko [7]2 years ago

5 0

I'm really not allowed to follow a link outside of Brainly. Don't make a habit of it. I take it we are still on y = 8*(1.2)^x

Rate of change = (y2 - y1)/(x2 - x1) = (13.824 - 9.6) / 3 - 1

Rate of change = (4.224) / 2 = 2.112

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Work out the shaded area

Dennis_Churaev [7]

Answer:

Step-by-step explanation:

Shaded area:

Area of big rectangle - Area of small rectangle

$= 10 \times 12 - (10 - 5) \times (12 - 2)$

Calculate the difference

$= 10 \times 12 - 5 \times 10$

Calculate the product

$= 120 - 50$

Calculate the difference

$= 70 \: {cm}^{2}$

Hope this helps...

Best regards!!

7 0

3 years ago

How many days is 1,800 minutes?

nikklg [1K]

ITS 1.25

YOUR WELCOME

8 0

2 years ago

A bicycle tire has a diameter of 3 feet. How many turns of the wheel could it take for the bicycle to go 792 feet?

Andrew [12]

Answer:

84 complete turns

Step-by-step explanation:

In order to find the number of turns that the bicycle wheel is going to take while going 792 feet, we need to find the circumference of the wheel.

$C = \pi \: d = 3.14 \times 3 = 9.42 \: ft \\ no \: of \: turns = \frac{792}{9.42} \\ = 84.0764331 \\ = 84 \: turns$

7 0

3 years ago

identify an equation in point-slope form for the line parallel to y=-2/3x+8 that passes through (4,-5)

MariettaO [177]

Answer:

y +5 = (-2/3)(x -4)

Step-by-step explanation:

The point-slope form of an equation of a line with slope m through point (h, k) is ...

... y -k = m(x -h)

You have (h, k) = (4, -5) and a slope the same as that of the parallel line, m = -2/3. Then the equation is ...

... y -(-5) = (-2/3)(x -4)

... y +5 = (-2/3)(x -4)

7 0

2 years ago

Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

3 0

2 years ago