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2 years ago

If AC = 14, what is the measure of AD? Round your answer to the nearest hundredth.

Mathematics

1 answer:

Vitek1552 [10]2 years ago

3 0

Answer:

If I could rearrange the alphabet, I’d put ‘U’ and ‘I’ together

Step-by-step explanation:

You might be interested in

What is the third derivative of tan x?

azamat

The First is sec^2(x). Second is 2sec^2tanx. From that you get 2[2sec^2xtan^x+sec^4x]

4 0

3 years ago

Out of 30 questions, Ahmad answer 12 of them incorrectly. What percent of the questions did he answer correctly?

kati45 [8]

First, you would need to determine how many questions he got correct.
30 - 12 = 18

Next, you would need to create an equation for this problem. Let x represent the unknown percentage.
$\frac{18}{30}$ = $\frac{x}{100}$

Now, you would cross multiply.
30x = 1,800

The last step would be to isolate the x. To do this, you would divide both sides of the equal sign by 30.
x = 60

Ahmad got 60% of the questions correct.

I hope this helps!

7 0

2 years ago

15. Explain why we need to specify 0 < b < 1 and b > 1 as valid values for the base b in the expression logb(xx).

damaskus [11]

Answer:

The base (b) has to be positive and different of 1. The logarithm is the inverse of exponential, so:

logb(a) = x ⇒ a = bˣ

So, for b = 0 ⇒ 0ˣ = a

And there is impossible, "a" only could be 0.

For b = 1 ⇒ 1ˣ = a

And the same thing would happen, the logarithming would be to be 1, and the function will be extremally restricted.

For b<0, then the expression a = bˣ will be also restricted, and will not represent all values of a.

So, 0<b<1 and b >1.

3 0

2 years ago

Find all solutions of the equation in the interval [0, 2pi).

natali 33 [55]

Answer:

Step-by-step explanation:

Begin by squaring both sides to get rid of the radical. Doing that gives you:

$sin^2x=1-cosx$

Now use the Pythagorean identity that says

$sin^2x =1-cos^2x$ and make the replacement:

$1-cos^2x=1-cosx$ . Now move everything over to one side of the equals sign and set it equal to 0 so you can factor:

$1-cos^2x+cosx-1=0$ and then simplify to

$cosx-cos^2x=0$

Factor out the common cos(x) to get

$cosx(1-cosx)=0$ and there you have your 2 trig equations:

cos(x) = 0 and 1 - cos(x) = 0

The first one is easy enough to solve. Look on the unit circle and see where, one time around, where the cos of an angle is equal to 0. That occurs at

$x=\frac{\pi }{2},\frac{3\pi}{2}$