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3 years ago

Part A

Mathematics

2 answers:

wlad13 [49]3 years ago

3 0

Answer:

Part 1:

A. Correct

B. Correct

Part 2:

A. Correct

C. Correct

nikitadnepr [17]3 years ago

3 0

Answer:

I got 100 % on the quiz

Step-by-step explanation:

check the photo below for the awnser

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Which number is divisible by 5 A 44 B 125 C 116 D 114

Ray Of Light [21]

Answer:

Step-by-step explanation:

7 0

4 years ago

Read 2 more answers

At 2:00 p.m. two cars start toward each other from towns that are 240 miles apart. The rate of one car is 15 mph faster than the

Vesnalui [34]

Hello! :)

The answer is x+15!

~Hope it helps!~ :)

3 0

3 years ago

Derivative, by first principle <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

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

7 0

4 years ago

A box of chocolates contains 18 chocolate squares, tens of the square are milk chocolate, 5 are dark chocolate, and 3 are.white

Step2247 [10]

Answer:

0.167 ; 0.093

Step-by-step explanation:

Given that:

Total number of chocolate squares = 18 of which÷

Number of milk chocolate square = 10

Dark chocolate square = 5

White chocolate square= 3

Probability of choosing a white chocolate square :

Probability = required outcome / Total possible outcomes

Required outcome = white chocolate square = 3

Number of chocolate square = 18

P(choosing a white chocolate square) = 3 / 18 = 1/6 =

B) probability of randomly selecting a milk chocolate first, replace it, and then select a white chocolate?

With replacement :

First pick :

Number milk chocolate = 10

P(choosing a milk chocolate) = 10/18 = 5/9

Second pick:

Number of white chocolate = 3

P(choosing a white chocolate) = 3 / 18 = 1/6

Hence, since both probabilities are independent :

P(choosing milk, then white chocolate) :

(5/9 × 1/6) = 5/54 = 0.0925 = 0.093

8 0

4 years ago

What is the average rate of change of the function over the interval x = 0 to x = 5?

AURORKA [14]

$\bf slope = m = \cfrac{rise}{run} \implies \cfrac{ f(x_2) - f(x_1)}{ x_2 - x_1}\impliedby \begin{array}{llll} average~rate\\ of~change \end{array}\\\\[-0.35em] \rule{34em}{0.25pt}\\\\ f(x)= 2x^2-1\qquad \begin{cases} x_1=0\\ x_2=5 \end{cases}\implies \cfrac{f(5)-f(0)}{5-0} \\\\\\ \cfrac{[2(5)^2-1]~~-~~[2(0)^2-1]}{5}\implies \cfrac{50-(-1)}{5}\implies \cfrac{50+1}{5}\implies \cfrac{51}{5}\implies 10\frac{1}{5}$

6 0

3 years ago