How do you solve this?

Solve and get brainliest!! please be sure of your answer!!!!

laila [671]

Answer:

hope this helps youuuuu

4 0

3 years ago

Simplify 2^-6 times 2^3

IgorC [24]

Answer:

(2^-6)*(2^3)

=2^(-6+3)

=2^-3

=1/(2^3)

=1/8

3 0

3 years ago

Read 2 more answers

Value of (–3/8) × (+8/15). A. 11/15 B. 11/23 C. –1/5 D. –1/3

svet-max [94.6K]

ANSWER

The correct answer is C

EXPLANATION

We want to find the value of

$- \frac{3}{8} \times \frac{8}{15}$

This is the same as:

$- \frac{3}{8} \times \frac{8}{3 \times 5}$

We cancel out the common factors to get:

$- \frac{1}{1} \times \frac{1}{1 \times 5}$

We simplify to get:

$- \frac{1}{5}$

The correct answer is C

5 0

3 years ago

Let us analyze the following setting. You are given a circle of unit circumference. You pickkpoints on the circle independently

garik1379 [7]

Answer:

We have been given a unit circle which is cut at k different points to produce k different arcs. Now we can see firstly that the sum of lengths of all k arks is equal to the circumference:

$\small \sum_{i = 1}^{k} L_i= 2\pi$

Now consider the largest arc to have length \small l . And we represent all the other arcs to be some constant times this length.

we get :

$\small \sum_{i = 1}^{k} C_i.l = 2\pi$

where C(i) is a constant coefficient obviously between 0 and 1.

$\small \sum_{i = 1}^{k} C_i= 2\pi/l$

All that I want to say by using this step is that after we choose the largest length (or any length for that matter) the other fractions appear according to the above summation constraint. [This step may even be avoided depending on how much precaution you wanna take when deriving a relation.]

So since there is no bias, and \small l may come out to be any value from [0 , 2π] with equal probability, the expected value is then defined as just the average value of all the samples.

We already know the sum so it is easy to compute the average :

$\small L_{Exp} = \frac{2\pi}{k}$

7 0

3 years ago

Complete the proof Prove: UVY = UZW A) AAS B) ASA C) SAS D) SSS

puteri [66]

Answer:

B) ASA

Step-by-step explanation:

The ASA (Angle-Side-Angle) postulate states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent. (The included side is the side between the vertices of the two angles)

In this case, angle U, side UY and angle Y are congruent to angle U, side UW and angle W

7 0

3 years ago