All categories

2 years ago

What is twice 2^30 Write using exponents

Mathematics

1 answer:

quester [9]2 years ago

8 0

$\bf 2\cdot 2^{30}\implies 2^1 \cdot 2^{30}\implies 2^{1+30}\implies 2^{31}$

You might be interested in

Consider the line represented by the equation 5x + 2y = 10. How is the slope of the line related to values of A, B, and C in sta

Flauer [41]

Step-by-step explanation:

5x + 2y = 10

Ax + By = C

A = 5

B = 2

C = 10

5 0

2 years ago

Help again quick please!

Dennis_Churaev [7]

Answer:

2/3

Step-by-step explanation:

5/6 - (1/8÷3/4)

=5/6 - (1/8 × 4/3)

=5/6 - ( 1/6)

=4/6

=2/3

8 0

2 years ago

Read 2 more answers

maze going to the lake to visit some friends at the lake is 60 miles away and Renee is driving 40 mph the entire time how long w

mojhsa [17]

Answer:

it will take her an hour and 25 minutes roughly

Step-by-step explanation:

7 0

2 years ago

Karen runs a print shop that makes posters for large companies. It is a very competitive business. The market price is currently

kumpel [21]

AFC = FC / Quantity printed

So given she prints 1,000 posters: AFC = 250.00/1000 = $0.25 
Given she prints 2,000 posters: AFC = 250.00/2000 = $0.125 
Given she prints 10,000 posters: AFC = 250.00/2000 = $0.025 

ATC = TC / Quantity printed 

where TC = FC + Variable C * Quantity printed 

If she prints 1000: TC = 250 + 2000*1000 = 2,000,250 
ATC = 2,000,250/1000 = 2000.25 

If she prints 2000: TC = 250 + 1600*2000 = 3,200,250 
ATC = 3,200,250/2000 = 1600.125 

If she prints 10000: TC = 250 + 1600*2000 + 1000*8000 ($1000 for each additional poster after 2000) = 11,200,250 
ATC = 11,200,250/10000 = 1120.025

7 0

2 years ago

Solve the system of equations. x+y=4 y=x^2 - 8x + 16 a) {(-3,7).(-4, 8)} b) [(4,0)} c) {(3,1),(4,0) d) {(3,7). (4.0)} e) none

Marysya12 [62]

Answer: The required solution of the given system is

(x, y) = (3, 1) and (4, 0).

Step-by-step explanation: We are given to solve the following system of equations :

$x+y=4~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\y=x^2-8x+16~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

From equation (i), we have

$x+y=4\\\\\Rightarrow y=4-x~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)$

Substituting the value of y from equation (iii) in equation (ii), we get

$y=x^2-8x+16\\\\\Rightarrow 4-x=x^2-8x+16\\\\\Rightarrow x^2-8x+16-4+x=0\\\\\Rightarrow x^2-7x+12=0\\\\\Rightarrow x^2-4x-3x+12=0\\\\\Rightarrow x(x-4)-3(x-4)=0\\\\\Rightarrow (x-3)(x-4)=0\\\\\Rightarrow x-3=0,~~~~~~~x-4=0\\\\\Rightarrow x=3,~4.$

When, x = 3, then from (iii), we get

$y=4-3=1.$

And, when x = 4, then from (iii), we get

$y=4-4=0.$

Thus, the required solution of the given system is

(x, y) = (3, 1) and (4, 0).

8 0

3 years ago