All categories

3 years ago

14.) If AB = 13.BC = 6,CD = 2x - 18. and AD = 4x - 29 find CD.

Mathematics

1 answer:

natka813 [3]3 years ago

5 0

Answer:

Step-by-step explanation:

13 + 6 + 2x -18 = 4x - 29

19 + 2x -18 = 4x -29

2x + 1 = 4x - 29

30 = 2x

15 = x

CD = 2(15) -18

CD = 30 -18

CD = 12

You might be interested in

we have 4 1/2 cups of chocolate chips. a batch of our madison cookies requires 3/4 cup of chocolate chips. how many batches can

My name is Ann [436]

Answer:

You can make 6 batches of cookies.

Step-by-step explanation:

3/4 + 3/4 = 1 1/2 (2)

1 1/2 + 1 1/2 = 3 (4)

3 + 1 1/2 = 4 1/2 (6)

7 0

3 years ago

HELP PLEASE! I would really appreciate it :)

dem82 [27]

Each time x increases by 2 (eg: from x = -3 to x = -1), the value of y increases by 7 (eg: y = 9.5 to y = 16.5)

Change in y = 7
change in x = 2
Slope = (change in y)/(change in x) = 7/2 = 3.5

This all means that we have a linear equation

4 0

3 years ago

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

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

7 0

3 years ago

Andrew wants to lay new turf in his garage.Andrews garden is a rectangle measuring 3m by 4m,New turf cost 12.50pr m2.

Mrac [35]

Answer:

$150

Step-by-step explanation:

The area of the lawn is 12m so 12 x 12.50 is 150. It would cost $150

7 0

2 years ago

I will mark brainlyest plss helpp

lana [24]

Answer:

The answer is C

Step-by-step explanation:

i also use khan

5 0

2 years ago

Read 2 more answers