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3 years ago

Match the solution set of each of the following inequalities with its graph.

Mathematics

1 answer:

8090 [49]3 years ago

7 0

Answer:

Step-by-step explanation:

From the picture attached,

Number line Inequality

1). Arrow starting with a solid point from x = -2 x ≤ -2

directing towards negative numbers of x 2).Arrow starting with a hollow circle from x = 2 x >2

directing towards positive numbers of x

3). Arrow starting with a hollow circle from x = 2 x < 2 ⇒ x - 2 < 0

directing towards negative numbers of x

4). Arrow starting with a solid circle from x = 2 x ≥ 2 Or 2 ≤ x

directing towards positive numbers of x

5). Arrow starting with a hollow circle from x = -2 x > -2

directing towards positive numbers of x

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Sharon’s house, the library, and Lisa’s house are all on the same straight road. Sharon has to ride her bike 1 3/5 miles to get

stellarik [79]

4.35 miles is the answer

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2 years ago

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A restaurant makes pizzas with 6-inch diameters ad 12-inch diameters. Which of the following statement is true?

a_sh-v [17]

Answer:

The correct option is c: the area of the 6-inch pizza is 1/4 the area of the 12-inch pizza.

Step-by-step explanation:

The area (A) of each pizza is given by:

$A = \pi r^{2}$

Where:

r: is the radius

For the 12-inch pizza we have:

$A_{12} = \pi (\frac{12 inch}{2})^{2} = 113.1 inch$

And for the 6-inch pizza we have:

$A_{6} = \pi (\frac{6 inch}{2})^{2} = 28.3 inch$

The ratio between A₆ and A₁₂ is:

$\frac{A_{6}}{A_{12}} = \frac{28.3 inch}{113.1 inch} = 0.25 = \frac{1}{4}$

Hence:

$A_{6} = \frac{A_{12}}{4}$

Therefore, the correct option is c: the area of the 6-inch pizza is 1/4 the area of the 12-inch pizza.

I hope it helps you!

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3 years ago

Complete the open array model to find 1,950 ÷ 6.

gizmo_the_mogwai [7]

Answer:

The answer is 325

Step-by-step explanation:

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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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3 years ago

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ollegr [7]

Mike is driving 20 miles per hour

If he is driving 10 miles in 30 mins
(1/2 an hour) if you times both by two it tells you the MPH that Mike is driving.

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3 years ago