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4 years ago

5

What is the average rate of change of the function f(x)= 2^x - 3 from x= 4 and x= 6?

1 answer:

WITCHER [35]4 years ago

7 0

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-3x is less than or equal to 3, the solution of the inequality is?

Charra [1.4K]

Answer:

you dont really want the smoke

Step-by-step explanation:

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3 years ago

Can someone help me find the equivalent expressions to the picture below? I’m having trouble

miss Akunina [59]

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is $\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$ .

Now we will solve this expression with the help of law of exponents.

$\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}$

$=2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}$

$=2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}$

$=2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}$

$=2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }$ [Option 2]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$ [Option 1]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$

$=(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }$

$=\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} }$ [Option 3]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }$

$=\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}}$ [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

6 0

3 years ago

Image Above.........

Savatey [412]

Answer:

4 : 39

5 : 52

6 : 67

7 : 77

Step-by-step explanation:

7 0

3 years ago

Johnathan ran 5 days this week. The most he ran in one day was 3.5 miles. Write an inequality that shows the distance johnathan

Mekhanik [1.2K]

<em><u>An inequality that shows the distance Johnathan could of ran any day this week is:</u></em>

$x\leq 3.5$

<em><u>Solution:</u></em>

Let "x" be the distance Johnathan can run any day of this week

Given that,

Johnathan ran 5 days this week. The most he ran in one day was 3.5 miles

Therefore,

Number of days ran = 5

The most he ran in 1 day = 3.5 miles

Thus, the maximum distance he ran in a week is given as:

$distance = 5 \times 3.5 = 17.5$

The maximum distance he ran in a week is 17.5 miles

If we let x be the distance he can run any day of this week then, we get a inequality as:

$x\leq 3.5$

If we let y be the total distance he can travel in a week then, we may express it as,

$y\leq 17.5$

8 0

3 years ago

Descried how to derive the quadratic formula from a quadratic equation in standard form

Aleks [24]

Answer:

The standard form of a quadratic equation is:

$ax^2+bx+c=0$ , $a\neq 0$

Quadratic Formula Derivation:

$ax^2+bx+c=0\\$

$x^2+\frac{b}{a}x+\frac{c}{a} =0$

$x^2+\frac{b}{a}x = -\frac{c}{a}$

Completing the Square:

$x^2+\frac{b}{a}x +\frac{b^2}{4a^2} = \frac{b^2}{4a^2}-\frac{c}{a}$

$( x+\frac{b}{2a} )^2 = \frac{b^2-4ac}{4a^2}$

Square Root property:

$x+\frac{b}{2a} = \pm \sqrt{ \frac{b^2-4ac}{4a^2}}$

$x = -\frac{b}{2a} \pm { \frac{\sqrt {b^2-4ac}}{2a}}$

$x = \frac{-b\pm\sqrt{b^2-4ac}} {2a} }$

8 0

3 years ago