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2 years ago

Find the missing value pls help

Mathematics

1 answer:

Georgia [21]2 years ago

3 0

Most probably “-5” cuz -5 -(-2) =-5+2 =-7

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Amira has 3/4 of a bag of cat food. Her cat eats 1/10 a bag per week. How many weeks will the food last?

Tema [17]

Answer:

trick question, the cat is going to die anyways so forget the food

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Which of the following represents 3x-5y+10=0 written in slope-intercept form?

taurus [48]

For this case we have that by definition, the equation of a line in the slope-intercept form is given by:

$y = mx + b$

Where:

m: Is the slope

b: Is the cut-off point with the y axis

We have the following equation:

$3x-5y + 10 = 0$

We manipulate algebraically:

We subtract 10 from both sides of the equation:

$3x-5y = -10$

We subtract 3x from both sides of the equation:

$-5y = -3x-10$

We multiply by -1 on both sides of the equation:

$5y = 3x + 10$

We divide between 5 on both sides of the equation:

$y = \frac {3} {5} x + \frac {10} {5}\\y = \frac {3} {5} x + 2$

Thus, the equation in the slope-intercept form is $y = \frac {3} {5} x + 2$

Answer:

$y = \frac {3} {5} x + 2$

5 0

3 years ago

BRAINLIEST ANSWER PLUS POINTS IM GOING TO BREAK MY COMPUTER PLS HELP!!! Complete the table of values then plot the ordered pairs

Anastasy [175]

First, that looks like Pearson and if it is, I'm so sorry for you.
Secondly, I think this answer is correct.

5 0

2 years ago

Roman completed the square to solve the equation 0=−x2+6x−15 0 = − x 2 + 6 x − 15 as follows. Step 1: x2−6x+15=0 x 2 − 6 x + 15

aleksandr82 [10.1K]

Answer:

yes

Step-by-step explanation:

at step 6, his meant to add 3 to both side not subtract

and at step 7, it's meant to be i√6 not i6

hope this helps:)

5 0

1 year ago

Write an equivalent series with the index of summation beginning at n = 1. ∞ (−1)n + 1(n + 1)xn n = 0

kkurt [141]

Looks like the starting series is

$\displaystyle\sum_{n=0}^\infty\left((-1)^n+1\right)(n+1)x^n$

Note the degree of the first few terms in the series, for $n=0,1,2$ , are $x^0,x^1,x^2$ , and so on.

So if the series were to start at $n=1$ , we would need to preserve that pattern, and we do that by replacing $n$ with $n-1$ :

$\displaystyle\sum_{n=1}^\infty\left((-1)^{n-1}+1\right)(n-1+1)x^{n-1}$

Simplifying a bit, we end up with

$\displaystyle\sum_{n=1}^\infty\left(1-(-1)^n\right)nx^{n-1}$

7 0

2 years ago