Please helppp I really need this As soon as possible !

Write a ratio and a percent for the shaded area.

alisha [4.7K]

HEYA MATE 

YOUR ANSWER IS A.3/10,30%

BECAUSE SHADED SQUARES ARE 6 

AND TOTAL SQUARES ARE 20

THAN APPLY THE FORMULA OF PERCENTAGE.

=>GIVEN NUMBER/TOTAL NUMBER×100
6/20×100

THAN WE GET

30%

THANK YOU 

4 0

3 years ago

How much do you have to add to 1,750 m to get 4 m?

Dmitrij [34]

Answer:

is the questions even correct? recheck the units

4 0

3 years ago

5. Write down the nth term of each of the following G.Ps

yawa3891 [41]

Answer:

1i) nth term = 12(-b/4)^(n-1)

ii) nth term = 3(-1/9)^(n-1)

2i) Sixth term= (-3b^5)/256

ii) Eight term = -1/1594323

Question:

The complete question as found on brainly (question-10746111):

Write down the nth term of each of the following G.Ps whose first two terms are given as follows. Also find the term stated besides each G.P.

i) 12,-3b,......sixth term

ii) 3,-1/3,..........;8th term?

Step-by-step explanation:

1) To solve terms involving Geometric progressions (GPs), we would first state the variables that are to be incorporated in the formula.

i) 12,-3b,......;sixth term

1st term = a= 12

r = common ratio = 2nd term /1st term

r = -3b/12 = -b/4

Then we would find the nth term

See attachment for details

ii) 3,-1/3,..........;8th term

1st term = a= 3

r = common ratio = 2nd term /1st term

r = (-⅓)/3 = (-⅓)(⅓) = -1/9

Then we would find the nth term

See attachment for details

2) we are to determine the sixth term in (i) and eight term in (ii)

See attachment for details

Sixth term= (-3b^5)/256

Eight term = -1/1594323

7 0

3 years ago

Solve for y. y + 1.05 = 7.26

sleet_krkn [62]

6.21 --subtract 1.05 from 7.26

5 0

3 years ago

Please Help!! These are 3 seperate questions I need help with.

Brums [2.3K]

Answer:

A. (x1, x2) = (1, 4)
C. There is no solution.
B. The solutions are of the form: (x1, x2, x3) = (4t+4, t-3, t)

Step-by-step explanation:

1. Subtract the first equation from twice the second to eliminate x2.

2(2x1 -x2) -(x1 -2x2) = 2(-2) -(-7)

3x1 = 3 . . . . . simplify

x1 = 1 . . . . . . .divide by 3

Substitute this value into the second equation.

2·1 -x2 = -2

4 -x2 = 0 . . . . . add 2

4 = x2 . . . . . . . add x2

The solution is (x1, x2) = (1, 4).

__

2. Multiply the first equation by 3 and add the second equation.

3(-5x1 -3x2) +(15x1 +9x2) = 3(7) +(2)

0 = 23 . . . . . not true;

There is no solution.

__

3. There are two equations in 3 unknowns, so there cannot be a unique solution. The equations are not dependent, so there will be an infinite number of solutions that can be written in terms of a single parameter (t).

Let x3 = t. Then the system of equations can be rewritten as

2x1 - 5x2 = 23 +3t

x1 - 4x2 = 16

Subtracting twice the second equation from the first, we have ...

(2x1 -5x2) -2(x1 -4x2) = (23 +3t) -2(16)

3x2 = 3t -9 . . . simplify

x2 = t -3 . . . . . divide by 3

Substituting this into the second equation above, we have ...

x1 -4(t -3) = 16

x1 = 16 +4t -12 . . . . . . add 4(t-3)

x1 = 4t +4 . . . . . . . . simplify

The solutions are (x1, x2, x3) = (4t+4, t-3, t).

_____

The attachment shows the first system solved by graphical means. This validates our answer.

7 0

3 years ago