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olga nikolaevna [1]

2 years ago

6

identify which number is greater with the symbols > or < or ~ (approximate) pls have the explanation as well thank you!! -

10/3 -3.3

1 answer:

grin007 [14]2 years ago

4 0

Answer:

-3.3>-10/3

Step-by-step explanation:

-10.3 is equal to -3.3333333333....

-3.3 just stops at the tenths so there is 0.033333333 more in -10/3. The higher the negative number the lower it is, for example -3>-4. Therefore, -3.3>-10/3.

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Erin buys a bag of peanuts that weighs 3/4 of a pound.later that week the bag is 2/3 full how much does the back of peanuts weig

nasty-shy [4]

The answer is 1/2lbs, here's how.

3/4x2/3=6/12 reduced to 1/2lbs.

4 0

2 years ago

Follow the process of completing the square to solve 2x2 + 8x - 12 = 0. After adding B2 to both sides of the equation in step 4,

Katena32 [7]

2x^2 + 8x - 12 = 0..divide by 2
x^2 + 4x - 6 = 0
x^2 + 4x = 6...add 4 to both sides of the equation
x^2 + 4x + 4 = 6 + 4
(x + 2)^2 = 10....<== ur constant is 10
x + 2 = (+-)sqrt 10
x = -2 (+ - ) sqrt 10

x = -2 + sqrt 10
x = -2 - sqrt 10

3 0

3 years ago

Read 2 more answers

Plsss hellpppp 14 and 15 plsssssss

Lesechka [4]

Answer:

14). 2nd quadrant

15). 1st quadrant

Step-by-step explanation:

14).Coordinates of a point → J(-8, -12)

Coordinates of the new point J' after reflection of x-axis will follow the rule,

(x, y) → (x, -y)

Coordinates of J' → (-8, 12)

Therefore, point J' will lie in 2nd quadrant.

15). Coordinates of a point → W(-6, 7)

Rule for the rotation by 90°clockwise about the origin,

(x, y) → (y, -x)

Coordinates of point W → (-6, 7)

Following this rule,

W(-6, 7) → W'(7, 6)

Therefore, point W' will lie in the first quadrant.

7 0

2 years ago

Help help help help help help help help help help

AysviL [449]

Answer:

1/4

GO GOOD LUCK

Step-by-step explanation:

U doing IXL? LOL

make me brainliest please

7 0

2 years ago

Read 2 more answers

Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

Bezzdna [24]

Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$

so the left side is

$2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n$

Also recall that

$\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}$

so that the right side is

$b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)$

Solve for $c$ .

$2n^2 + 2(c-1)n = c(2^n-1) \implies c = \dfrac{2n^2 - 2n}{2^n - 2n - 1} = \dfrac{2n(n-1)}{2^n - 2n - 1}$

Now, the numerator increases more slowly than the denominator, since

$\dfrac{d}{dn}(2n(n-1)) = 4n - 2$

$\dfrac{d}{dn} (2^n-2n-1) = \ln(2)\cdot2^n - 2$

and for $n\ge5$ ,

$2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2$

This means we only need to check if the claim is true for any $n\in\{1,2,3,4\}$ .

$n=1$ doesn't work, since that makes $c=0$ .

If $n=2$ , then

$c = \dfrac{4}{2^2 - 4 - 1} = \dfrac4{-1} = -4 < 0$

If $n=3$ , then

$c = \dfrac{12}{2^3 - 6 - 1} = 12$

If $n=4$ , then

$c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N$

There is only one value for which the claim is true, $c=12$ .

3 0

1 year ago