All categories

2 years ago

I need help please and can you explain ANSWER ASAP!!!!

Mathematics

1 answer:

Andru [333]2 years ago

5 0

The app uses 2^28 bytes.

Step-by-step explanation:

Step 1: Given total storage used by the app = 4^4 Megabytes. Also, 1 MB = 2^20 bytes. Find total storage used by app in bytes.

⇒ 4^4 × 2^20 = (2²)^4 × 2^20 = 2^8 × 2^20

= 2^8+20

= 2^28 bytes (using the law of exponents a^m × a^n = a^m+n)

You might be interested in

Use the formula C= dxd

zysi [14]

Answer:

69.08m

Step-by-step explanation:

Circumference of a circle = πd where

d is the diameter of the circle

Given

diameter = 22metres

circumference of the circle = 3.14(22)

circumference of the circle = 69.08m

Hence the circumference of the circle is 69.08m

5 0

2 years ago

Read 2 more answers

What Is the volume ?

MrMuchimi

336......?? find the area of cross section and than multiply with the lenght

5 0

3 years ago

Read 2 more answers

Select the correct answer,<br> How does the figure verify the triangle inequality theorem?

skelet666 [1.2K]

Step-by-step explanation:

just think it through. it can only be the first option. the sides 7 and 4 are together 11, and they can only meet, when putting them flat on the 11 side. but yes, if you do that, they will meet, as 11 = 11.

the result would be an extreme triangle : appearing just as one, flat line (length 11), the angles being 180°, 0° and 0°.

6 0

2 years ago

wo point charges lie on the $x$ axis. A charge of +6.24 $\mu C$ is at the origin, and a charge of -9.55 $\mu C$ is at $x$ = 12.0

Andreyy89

Answer:

$E_n=34,467,075.42\ N/C$

Step-by-step explanation:

<u>Electric Field</u>

The electric field produced by a point charge Q at a distance d is given by

$\displaystyle E=K\cdot \frac{Q}{d^2}$

Where

$K = 9\cdot 10^9\ Nw.m^2/c^2$

The net electric field is the vector addition of the individual electric fields produced by each charge. The direction is given by the rule: If the charge is positive, the electric field points outward, if negative, it points inward.

Let's calculate the electric fields of each charge at the given point. The first charge $q_1=+6.24\mu C=6.24\cdot 10^{-6}C$ is at the origin. We'll calculate its electric field at the point x=-3.85 cm. The distance between the charge and the point is d=3.85 cm = 0.0385 m, and the electric field points to the left:

$\displaystyle E_1=9\cdot 10^9\cdot \frac{6.24\cdot 10^{-6}}{0.0385^2}$

$E_1=37,888,345.42\ N/C$

Similarly, for $q_2=-9.55\mu C=-9.55\cdot 10^{-6}C$ , the distance to the point is 12 cm + 3.85 cm = 15.85 cm = 0.1585 m. The electric field points to the right: