Determine if ΔABC and ΔFHG are similar. If so, write the similarity statement.options:


A) The triangles are not similar



B) ΔABC ∼ ΔHFG


C) Impossible to determine.


D) ΔABC ∼ ΔFHG

Answer:

17

Step-by-step explanation:

Would you mind giving me brainliest?

Answer:

5 feet per 1 cm

Step-by-step explanation:

Answer:

The answer is 5 feet.

Step-by-step explanation:

Set up a proportion.

3/15 = 1/x (x being the number of feet)

Cross multiply.

3x = 15

Solve for x. (divide both sides by 3)

x = 5

I hope this helps! :)

Answer:

19.8 or 19 4/5

Step-by-step explanation:

Step 1: Set up the expression.

Step 2: Simplify.

Therefore, the answer is 19.8 or 19 4/5.

Answer:

0.5

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X higher than x is given by the following formula.

The mail arrival time to a department has a uniform distribution over 5 to 45 minutes.

This means that .

What is the probability that the mail arrival time is more than 25 minutes on a given day?

So the probability that the mail arrival time is more than 25 minutes on a given day is 0.5.

Answer:

(a) ⅛ tan⁻¹(¼)

(b) sec x − ln│csc x + cot x│+ C

Step-by-step explanation:

(a) ∫₀¹ x / (16 + x⁴) dx

∫₀¹ (x/16) / (1 + (x⁴/16)) dx

⅛ ∫₀¹ (x/2) / (1 + (x²/4)²) dx

If tan u = x²/4, then sec²u du = x/2 dx

⅛ ∫ sec²u / (1 + tan²u) du

⅛ ∫ du

⅛ u + C

⅛ tan⁻¹(x²/4) + C

Evaluate from x=0 to x=1.

⅛ tan⁻¹(1²/4) − ⅛ tan⁻¹(0²/4)

⅛ tan⁻¹(¼)

(b) ∫ (sec³x / tan x) dx

Multiply by cos x / cos x.

∫ (sec²x / sin x) dx

Pythagorean identity.

∫ ((tan²x + 1) / sin x) dx

Divide.

∫ (tan x sec x + csc x) dx

Split the integral

∫ tan x sec x dx + ∫ csc x dx

Multiply second integral by (csc x + cot x) / (csc x + cot x).

∫ tan x sec x dx + ∫ csc x (csc x + cot x) / (csc x + cot x) dx

Integrate.

sec x − ln│csc x + cot x│+ C

Answer:

(a) Solution : 1/8 cot⁻¹(4) or 1/8 tan⁻¹(¼) (either works)

(b) Solution : tan(x)/sin(x) + In | tan(x/2) | + C

Step-by-step explanation:

(a) We have the integral (x/16 + x⁴)dx on the interval [0 to 1].

For the integrand x/6 + x⁴, simply pose u = x², and du = 2xdx, and substitute:

1/2 ∫ (1/u² + 16)du

'Now pose u as 4v, and substitute though integral substitution. First remember that we have to factor 16 from the denominator, to get 1/2 ∫ 1/(16(u²/16 + 1))' :

∫ 1/4(v² + 1)dv

'Use the common integral ∫ (1/v² + 1)dv = arctan(v), and substitute back v = u/4 to get our solution' :

1/4arctan(u/4) + C

=> Solution : 1/8 cot⁻¹(4) or 1/8 tan⁻¹(¼)

(b) We have the integral ∫ sec³(x)/tan(x)dx, which we are asked to evaluate. Let's start by substitution tan(x) as sin(x)/cos(x), if you remember this property. And sec(x) = 1/cos(x) :

∫ (1/cos(x))³/(sin(x)/cos(x))dx

If we cancel out certain parts we receive the simplified expression:

∫ 1/cos²(x)sin(x)dx

Remember that sec(x) = 1/cos(x):

∫ sec²(x)/sin(x)dx

Now let's start out integration. It would be as follows:

Solution: tan(x)/sin(x) + In | tan(x/2) | + C

Answer:

Here we have the case where:

"one chip is eaten in 1/4 mins."

(that means unit rate, the amount of a given variable that we need to do a unit of another variable)

Then the unit rate would be:

4 chips per minute.

Or, in a more mathematical way to write this:

4 chips/min.

Now we could write this in other units, for example.

We know that 1 min = 60s.

Then we have that (1/4) min = 60s/4 = 15s.

Then we have the unit rate:

1 chip every 15 seconds, or:

(1/15) chips/second.

Now, the variable in this case is time, t  = time, let's multiply bot our unit rates by the same amount of time, and see if we have the same outcome.

Let's use t = 2 min.

1) (4chips/min)*2min = 8 chips.

2) ((1/15) chips/second)*2min

First, 2 min = 120 seconds.

((1/15) chips/second)*2min = ((1/15) chips/second)*120seconds = 8 chips.

So both unit rates are equivalent.