Graph the line with slope 3 and y-intercept -8.

Dalia saved $3000 last year. She said $600 in the month of January. What percent of the total amount of money she saved last yea

Anni [7]

1. We assume, that the number 3000 is 100% - because it's the output value of the task.
2. We assume, that x is the value we are looking for. 
3. If 100% equals 3000, so we can write it down as 100%=3000. 
4. We know, that x% equals 600 of the output value, so we can write it down as x%=600. 
5. Now we have two simple equations:
1) 100%=3000
2) x%=600
where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that:
100%/x%=3000/600
6. Now we just have to solve the simple equation, and we will get the solution we are looking for.

7. Solution for 600 is what percent of 3000

100%/x%=3000/600
(100/x)*x=(3000/600)*x - we multiply both sides of the equation by x
100=5*x - we divide both sides of the equation by (5) to get x
100/5=x 
20=x 
x=20
So 600 is 20% of 3000

7 0

3 years ago

6.Find the equation of the line: through: (-2, 3) and (1,5)

prohojiy [21]

Answer:

y=2/3x+4.3

Step-by-step explanation:

y=mx+c

(5-3)/(1--2)

2/3= gradient

y=2/3x +c

5=2/3(1)+c

-2/3

4.3=c

6 0

3 years ago

What is equivalent to 24b-34ab

jarptica [38.1K]

This is what I got as the answer, 2b x (12-17a)

6 0

3 years ago

A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021

egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

A is approximately 2020.022

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

$\displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}$

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

$\displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}$

Therefore, for the last term we have;

$\displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}$

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

$\displaystyle \frac{A}{2} = \mathbf{ 1011 \cdot \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460} \right)}$

$\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2} +\frac{1}{2} - \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022} \right)$