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3 years ago

How do you do this problem?

Mathematics

1 answer:

PIT_PIT [208]3 years ago

8 0

Answer:

Step-by-step explanation:

Let f(x) = 2x – 1

Substitute (x + b) in place of x.

f(x + b) = 2(x + b) – 1

= 2x + 2x –1

To find $\frac{f(x+b)-f(x)}{(x+b)-x}$ :

$\frac{f(x+b)-f(x)}{(x+b)-x}=\frac{(2x+2b-1)-(2x-1)}{(x+b)-x}$

$=\frac{2x+2b-1-2x+1}{x+b-x}$

$=\frac{2b}{b}$ (Cancel common term)

$=2$

Hence, $\frac{f(x+b)-f(x)}{(x+b)-x}=2.$

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Suppose that the proportions of blood phenotypes in a particular population are as follows: A B AB O 0.48 0.13 0.03 0.36 Assumin

Serggg [28]

Answer:

P(O and O) =0.1296

P=0.3778

Step-by-step explanation:

Given that

blood phenotypes in a particular population

A=0.48

B=0.13

AB=0.03

O=0.36

As we know that when A and B both are independent that

P(A and B)= P(A) X P(B)

The probability that both phenotypes O are in independent:

P(O and O)= P(O) X P(O)

P(O and O)= 0.36 X 0.36 =0.1296

P(O and O) =0.1296

The probability that the phenotypes of two randomly selected individuals match:

Here four case are possible

P=P(A and A)+P(B and B)+P(AB and AB)+P(O and O)

P=0.48 x 0.48 + 0.13 x 0.13 + 0.03 x 0.03 + 0.36 x 0.36

P=0.3778

7 0

3 years ago

If a coin is tossed 5 times, and then a standard six-sided die is rolled 4 times, and finally a group of three cards are drawn f

Step2247 [10]

Answer: 5,499,187,200

<u>Step-by-step explanation:</u>

A coin is tossed 5 times.

There are two options (heads or tails) so the possible outcomes are: 2⁵

A six-sided die is rolled 4 times.

There are six options so the possible outcomes are: 6⁴

A group of 3 cards are drawn (without replacement).

The first outcome has 52 options, the second has 51 options, and the third has 50 options: 52 x 51 x 50

Now if we want the coin AND the die AND the cards, we have to multiply all of their possible outcomes:

2⁵ x 6⁴ x 52 x 51 x 50

= 32 x 1296 x 132,600

= 5,499,187,200

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4 years ago

HELP QUICK will mark brainliest

slamgirl [31]

Answer:

Step-by-step explanation:

(5x4) - 6 / 2 = x

7 0

3 years ago

Read 2 more answers

Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

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4 years ago

Which expression is equivalent to 10x - (8x +3)

Verdich [7]

Answer:

you forgot to add the options of the answer

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3 years ago